Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 79



Work Step by Step

$ g(x)=\sqrt{x}$ $ g'(x)=\frac{1}{2\sqrt{x}}$ and $ g'(1)=\frac{1}{2}$ $f(u)=u^5+1$ $ f'(u)=5u^4$ $ f'(g(1))=f'(1)=5$ Therefore $(f o g)'(-1)=f'(g(1))g'(1)=5*\frac{1}{2}=\frac{5}{2}$
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