#### Answer

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

#### Work Step by Step

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

Published by
Cengage

ISBN 10:
1285740629

ISBN 13:
978-1-28574-062-1

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

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