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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