Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Review - Exercises - Page 271: 77

Answer

$$f'(x)=2g(x)g'(x)$$

Work Step by Step

$f'(x)=\frac{d}{dx}[g(x)]^2$ Using the chain rule: $f'(x)=\frac{d[g(x)]^2}{dg(x)}\times\frac{dg(x)}{dx}$ $=2g(x)\times g'(x)$ $=2g(x)g'(x)$
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