Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 7

Answer

$F'(x)=4(5x^{6}+2x^{3})^{3}(30x^{5}+6x^{2})$

Work Step by Step

$F(x)=(5x^{6}+2x^{3})^{4}$ Differentiate using the chain rule, which is $\dfrac{df(u)}{dx}=\dfrac{df}{du}\dfrac{du}{dx}$: $F'(x)=4(5x^{6}+2x^{3})^{3}(5x^{6}+2x^{3})'=...$ $...=4(5x^{6}+2x^{3})^{3}(30x^{5}+6x^{2})$
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