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How do you differentiate #f(x) = 5(x^2-4 )^(2) # using the chain rule?

Answer 1

Aurora Alvarado

#f'(x)=20x(x^2-4)#

#f(x)=5(x^2-4)^2# #f'(x)=5*2(x^2-4)*2x# #f'(x)=20x(x^2-4)#

The chain rule is given by:

#F(x)=f(g(x))# #F'(x)=f'(g(x))timesg'(x)#

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Answer 2

Isaiah Allen

To differentiate ( f(x) = 5(x^2-4)^2 ) using the chain rule, follow these steps:

Identify the outer function, which is ( u^2 ), where ( u = x^2 - 4 ).
Differentiate the outer function with respect to ( u ), which yields ( \frac{d}{du}(u^2) = 2u ).
Identify the inner function, ( u = x^2 - 4 ).
Differentiate the inner function with respect to ( x ), which yields ( \frac{d}{dx}(x^2 - 4) = 2x ).
Combine the results from steps 2 and 4 using the chain rule: ( \frac{df}{dx} = \frac{d}{du}(u^2) \cdot \frac{d}{dx}(x^2 - 4) = 2u \cdot 2x ).
Substitute back ( u = x^2 - 4 ) into the expression: ( \frac{df}{dx} = 2(x^2 - 4) \cdot 2x = 4x(x^2 - 4) ).

Therefore, the derivative of ( f(x) = 5(x^2-4)^2 ) with respect to ( x ) using the chain rule is ( \frac{df}{dx} = 4x(x^2 - 4) ).

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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.