## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6 - Page 158: 11

#### Answer

$f'( x ) =\frac{-12 (6x-2)}{(3x^2-2x+1)^4}$

#### Work Step by Step

Given that $f(x)=\frac{4}{(3x^2-2x+1)^3}$ We can first write it as $f(x)=4(3x^2-2x+1)^{-3}$ so that it is easier to deal with. Next we write this as $f(x)=4(g(x))^{−3}$, where $f(x)$ is the outside function and $g(x)=3x^2-2x+1$ is the inside function. To find the answer we need to differentiate the outside function $f(x)$ and times it by the derivative of the inside function $g(x)$. $f(x)=4(3x^2-2x+1)^{-3}$ $f (x)=4(g (x))^{-3}$ $f'(x)=-12(g ( x ))^{-3-1}×(g' ( x ))$ $f'(x)=-12(g ( x ))^{-4}×(6x^{2-1}-2)$ $f'(x)=-12(3x^2-2x+1)^{-4}×(6x^1-2)$ $f'(x)=-12(6x-2)(3x^2-2x+1)^{-4}$ $f'( x ) =\frac{-12 (6x-2)}{(3x^2-2x+1)^4}$

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