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: 10

Answer

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

Work Step by Step

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