## Calculus 10th Edition

Published by Brooks Cole

# Chapter 2 - Differentiation - 2.4 Exercises: 23

#### Answer

$f'(x)=(x)(x-2)^3(6x-4).$

#### Work Step by Step

Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=x^2 ;u’(x)=2x$ $v(x)=(x-2)^4$ $v'(x)$ is found using the Chain Rule: $u=(x-2)$; $\dfrac{du}{dx}=1$ $\dfrac{d}{du}v(u)=4u^3$ $\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=4(x-2)^3$ $f'(x)=2x(x-2)^4+4(x-2)^3(x^2)$ $=(x-2)^3(2x(x-2)+4x^2)$ $(x)(x-2)^3(6x-4).$

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