How do you derive #y = (x-1)/( x+3) ^ (2)# using the quotient rule?
Applying this rule to the given quotient of 2 functions yields :
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To derive ( y = \frac{(To derive ( y = \frac{(x-To derive ( y = \frac{(x-1To derive ( y = \frac{(x-1)}To derive ( y = \frac{{To derive ( y = \frac{(x-1)}{(To derive ( y = \frac{{xTo derive ( y = \frac{(x-1)}{(xTo derive ( y = \frac{{x -To derive ( y = \frac{(x-1)}{(x+3To derive ( y = \frac{{x - To derive ( y = \frac{(x-1)}{(x+3)^To derive ( y = \frac{{x - 1To derive ( y = \frac{(x-1)}{(x+3)^2} \To derive ( y = \frac{{x - 1}}{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using theTo derive ( y = \frac{{x - 1}}{{(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotientTo derive ( y = \frac{{x - 1}}{{(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule,To derive ( y = \frac{{x - 1}}{{(x +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate theTo derive ( y = \frac{{x - 1}}{{(x + To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator andTo derive ( y = \frac{{x - 1}}{{(x + 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominatorTo derive ( y = \frac{{x - 1}}{{(x + 3)^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separatelyTo derive ( y = \frac{{x - 1}}{{(x + 3)^2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately andTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and applyTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply theTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} )To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotientTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) usingTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient ruleTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using theTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formulaTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotientTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula,To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient ruleTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, whichTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule,To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which statesTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, weTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
IfTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiateTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate theTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( yTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator andTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y =To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominatorTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately,To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, thenTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{uTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then applyTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient ruleTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{vTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, whichTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states thatTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ),To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that theTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), thenTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivativeTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative ofTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( yTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y'To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' =To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \fracTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{fTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{uTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{gTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'vTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v -To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uvTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} )To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) isTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{vTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( uTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{fTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) andTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( vTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v ): To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). HereTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v ): To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's theTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v ): (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the stepTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v ): ( u =To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-byTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v ): ( u = (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-stepTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v ): ( u = (xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step processTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v ): ( u = (x -To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v ): ( u = (x - To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
1To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v ): ( u = (x - 1To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
1.To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1)To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
DifferentiateTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate theTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) )To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numeratorTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( fTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numeratorTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x)To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x -To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( vTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v =To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 )To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x +To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) toTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to getTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 )To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ). To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ). 2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
- Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate theTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' =To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x)To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain ruleTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative ofTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule:To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( gTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x -To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( vTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x)To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v'To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' =To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x +To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1)To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3)To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) )To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient ruleTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x +To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formulaTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula:To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( yTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y'To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' =To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \fracTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
3.To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
ApplyTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{fTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply theTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotientTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient ruleTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x)To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
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Apply the quotient rule: To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdotTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot gTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( yTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x)To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) -To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \fracTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - fTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x)To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdotTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
SimplTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot gTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( yTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y'To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' =To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(gTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \fracTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x +To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
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Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
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Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
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Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
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Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
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Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
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Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
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Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
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Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
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Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
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Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
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Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
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Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
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Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
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Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). 4To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 -To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
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Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). 4.To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
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SubstituteTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute theTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivativesTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x -To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives andTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functionsTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions intoTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into theTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formulaTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula:To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( yTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y'To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' =To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x +To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 -To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4}To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x -To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x +To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1)To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdotTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x -To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x -To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
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Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
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Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
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Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
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Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
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Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
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Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
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Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
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Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
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Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
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Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
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Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
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Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4}To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ). To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
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Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ). 5To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ). 5.To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- SimplTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
SimplifyTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expressionTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression:To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( yTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y'To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 -To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' =To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \fracTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 +To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x -To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
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Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
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Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
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Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
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Simplify the expression: ( y' = \frac{{(x + 3)^2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
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Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
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Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
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Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 -To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x -To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x +To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4}To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( yTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y'To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' =To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplifyTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y'To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' =To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \fracTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( yTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y'To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x -To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x -To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x +To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x +To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4}To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
-
Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
-
Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
-
Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
-
Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
-
Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
-
Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ). To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ). 7.To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So,To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, theTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifyingTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative ofTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: (To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( yTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( yTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y'To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y =To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' =To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \fracTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} )To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) isTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is (To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( yTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y'To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' =To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x -To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \fracTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-xTo derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}{{To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-x^2To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}{{(x +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-x^2 + 2x + 15}{(x + To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}{{(x + 3)^4}} ).
- Simplify the numeratorTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-x^2 + 2x + 15}{(x + 3To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}{{(x + 3)^4}} ).
- Simplify the numerator: ( y' = \frac{{(To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}{{(x + 3)^4}} ).
- Simplify the numerator: ( y' = \frac{{(xTo derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4}To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}{{(x + 3)^4}} ).
- Simplify the numerator: ( y' = \frac{{(x^To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} \To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}{{(x + 3)^4}} ).
- Simplify the numerator: ( y' = \frac{{(x^2To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
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Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
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Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} ).To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}{{(x + 3)^4}} ).
- Simplify the numerator: ( y' = \frac{{(x^2 +To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} ).To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}{{(x + 3)^4}} ).
- Simplify the numerator: ( y' = \frac{{(x^2 + 6To derive ( y = \frac{(x-1)}{(x+3)^2} ) using the quotient rule, you differentiate the numerator and denominator separately and apply the quotient rule formula, which states:
If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ).
Here's the derivation:
-
Identify ( u ) and ( v ): ( u = (x - 1) ) (numerator) ( v = (x + 3)^2 ) (denominator)
-
Differentiate ( u ) and ( v ): ( u' = 1 ) (derivative of ( x - 1 )) ( v' = 2(x + 3) ) (using the chain rule for ( (x + 3)^2 ))
-
Apply the quotient rule: ( y' = \frac{(1)(x + 3)^2 - (x - 1)(2(x + 3))}{(x + 3)^4} )
-
Simplify the expression: ( y' = \frac{(x + 3)^2 - 2(x - 1)(x + 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 3x - x - 3)}{(x + 3)^4} )
( y' = \frac{(x + 3)^2 - 2(x^2 + 2x - 3)}{(x + 3)^4} )
( y' = \frac{x^2 + 6x + 9 - 2x^2 - 4x + 6}{(x + 3)^4} )
( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} )
So, the derivative of ( y = \frac{(x-1)}{(x+3)^2} ) is ( y' = \frac{-x^2 + 2x + 15}{(x + 3)^4} ).To derive ( y = \frac{{x - 1}}{{(x + 3)^2}} ) using the quotient rule, we differentiate the numerator and denominator separately, then apply the quotient rule formula, which states that the derivative of ( \frac{{f(x)}}{{g(x)}} ) is ( \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ). Here's the step-by-step process:
- Differentiate the numerator ( f(x) = x - 1 ) to get ( f'(x) = 1 ).
- Differentiate the denominator ( g(x) = (x + 3)^2 ) using the chain rule: ( g'(x) = 2(x + 3)(1) = 2(x + 3) ).
- Apply the quotient rule formula: ( y' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}{{(g(x))^2}} ).
- Substitute the derivatives and functions into the formula: ( y' = \frac{{1 \cdot (x + 3)^2 - (x - 1) \cdot 2(x + 3)}}{{(x + 3)^4}} ).
- Simplify the expression: ( y' = \frac{{(x + 3)^2 - 2(x - 1)(x + 3)}}{{(x + 3)^4}} ).
- Further simplify: ( y' = \frac{{(x + 3)^2 - 2(x^2 + 2x - x - 3)}}{{(x + 3)^4}} ).
- Continue simplifying: ( y' = \frac{{(x + 3)^2 - 2(x^2 + x - 3)}}{{(x + 3)^4}} ).
- Simplify the numerator: ( y' = \frac{{(x^2 + 6x + 9) - 2x^2 - 2x + 6}}{{(x + 3)^4}} ).
- Combine like terms: ( y' = \frac{{-x^2 + 4x + 3}}{{(x + 3)^4}} ).
Therefore, the derivative of ( y = \frac{{x - 1}}{{(x + 3)^2}} ) is ( y' = \frac{{-x^2 + 4x + 3}}{{(x + 3)^4}} ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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