How do you differentiate #f(x) = (x)/(x^4-x^3+6)# using the quotient rule?

Answer 1

#f'(x)=(-3x^4+2x^3+6)/(x^4-x^3+6)^2#

The quotient rule states that

#d/dx[(g(x))/(h(x))]=(g'(x)h(x)-g(x)h'(x))/[h(x)]^2#

Applying this to the given function, we see that

#f'(x)=((x^4-x^3+6)d/dx[x]-xd/dx[x^4-x^3+6])/(x^4-x^3+6)^2#

Both of these derivatives can be found through the power rule:

#f'(x)=((x^4-x^3+6)(1)-(x)(4x^3-3x^2))/(x^4-x^3+6)^2#

Simplified, this gives

#f'(x)=(x^4-x^3+6-4x^4+3x^3)/(x^4-x^3+6)^2#
#f'(x)=(-3x^4+2x^3+6)/(x^4-x^3+6)^2#
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Answer 2

To differentiate ( f(x) = \frac{x}{x^4 - x^3 + 6} ) using the quotient rule, you apply the formula: ( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} ). Here, ( u = x ) and ( v = x^4 - x^3 + 6 ). Taking the derivatives, ( u' = 1 ) and ( v' = 4x^3 - 3x^2 ). Plugging these into the quotient rule formula, you get:

[ f'(x) = \frac{(x^4 - x^3 + 6) \cdot 1 - x \cdot (4x^3 - 3x^2)}{(x^4 - x^3 + 6)^2} ]

Simplify the expression:

[ f'(x) = \frac{x^4 - x^3 + 6 - (4x^4 - 3x^3)}{(x^4 - x^3 + 6)^2} ] [ f'(x) = \frac{-3x^4 + 4x^3 + 6}{(x^4 - x^3 + 6)^2} ]

That's the derivative of ( f(x) ) using the quotient rule.

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Answer from HIX Tutor

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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