How do you find the first and second derivatives of #y= (x^2 + 2x + 5) / (x + 1)# using the quotient rule?
First derivative:
Second derivative:
The rule for quotients is:
Applying the guideline:
That is the first derivative. Use this new fraction along with the same steps to find the next one:
After extensive simplification:
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first derivative:
second derivative:
Rule of the quotient:
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To find the first derivative of ( y = \frac{x^2 + 2x + 5}{x + 1} ) using the quotient rule, we differentiate the numerator and denominator separately and then apply the quotient rule formula ( \frac{d(u/v)}{dx} = \frac{v \cdot du/dx  u \cdot dv/dx}{v^2} ).

Differentiate the numerator: ( \frac{d}{dx}(x^2 + 2x + 5) = 2x + 2 )

Differentiate the denominator: ( \frac{d}{dx}(x + 1) = 1 )

Apply the quotient rule: ( \frac{d}{dx}\left(\frac{x^2 + 2x + 5}{x + 1}\right) = \frac{(x + 1)(2x + 2)  (x^2 + 2x + 5)(1)}{(x + 1)^2} )

Simplify the expression: ( \frac{(2x^2 + 2x + 2x + 2)  (x^2 + 2x + 5)}{(x + 1)^2} ) ( = \frac{x^2 + 4x + 2  x^2  2x  5}{(x + 1)^2} ) ( = \frac{2x  3}{(x + 1)^2} )
Therefore, the first derivative of ( y = \frac{x^2 + 2x + 5}{x + 1} ) is ( \frac{2x  3}{(x + 1)^2} ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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