Answer
$g'(x)=-108(4-9x)^3$
Work Step by Step
$u=4-9x$; $\dfrac{du}{dx}=-9$
$g(u)= 3u^4;\dfrac{d}{du}g(u)=12u^3$
$\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=12(4-9x)^3(-9)=-108(4-9x)^3.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 2 - Differentiation - 2.4 Exercises - Page 136: 9
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