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: 34

Answer

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

Work Step by Step

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