Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6 - Page 158: 7


$f'(x)=$ $37(x^3 + 2x)^{36}(3x^2+2) $

Work Step by Step

Given that $f(x)=(x^3+2x)^{37} $ We can write this as $ f (x)=(g (x))^{37} $, where $ f (x)$ is the outside function and $ g (x)=x^3+2x $, the inside function. Thus the answer is the derivative of the outside function, using the chain rule, times the derivative of the inside function. $ f (x)=(g (x))^{37}$ $f'(x)=37 (g (x))^{37-1}×(g'(x)) $ $f'(x)=37 (g (x))^{36}×(3x^{3-1}+2) $ $f'(x)=37 (x^3+2x)^{36}×(3x^2+2) $ $f'(x)=37 (x^3+2x)^{36}(3x^2+2) $
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