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: 33

Answer

$$y'=\frac{x}{\sqrt{1+x^2}}\cos\sqrt{1+x^2}$$

Work Step by Step

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