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: 10

Answer

$g'(x) = \frac{d}{dx} (2 - sin x)^{3/2} = \frac{-3cosx}{2}\sqrt {2-sinx}$

Work Step by Step

$g'(x) = \frac{d}{dx} (2 - sin x)^{3/2} = \frac{3}{2} (2 - sinx)^{1/2} * (0-cosx) = \frac{-3cosx}{2}\sqrt {2-sinx}$
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.