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: 33



Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.