Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises - Page 158: 34



Work Step by Step

$$y'=(\sqrt{\sin(1+x^2})'=\frac{1}{2\sqrt{\sin(1+x^2)}}(\sin(1+x^2))'=\frac{1}{2\sqrt{\sin(1+x^2)}}\cos(1+x^2)(1+x^2)'= \frac{1}{2\sqrt{\sin(1+x^2)}}\cos(1+x^2)\cdot2x= \frac{x\cos(1+x^2)}{\sqrt{\sin(1+x^2)}}$$
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