Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises: 24

Answer

$f'(x)=(2x-5)^2(8x-5)$

Work Step by Step

Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=x ;u’(x)=1 $ $v(x)=(2x-5)^3 $ $v'(x)$ is found using the Chain Rule: $u=(2x-5)$; $\dfrac{du}{dx}=2$ $\dfrac{d}{du}v(u)=3u^2$ $\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=2(3(2x-5)^2)=6(2x-5)^2.$ $f'(x)=(1)(2x-5)^3+(6(2x-5)^2)(x)=(2x-5)^2(8x-5).$
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