Answer
$h'(x) = \frac{1}{2}\sqrt{\frac{g(x)}{f(x)}}\cdot \frac{f'(x)~g(x)-f(x)~g'(x)}{[g(x)]^2}$
Work Step by Step
$h(x) = \sqrt{\frac{f(x)}{g(x)}}$
$h'(x) = \frac{1}{2}\Big(\frac{f(x)}{g(x)}\Big)^{-1/2}\cdot \frac{d}{dx}[\frac{f(x)}{g(x)}]$
$h'(x) = \frac{1}{2}\sqrt{\frac{g(x)}{f(x)}}\cdot \frac{f'(x)~g(x)-f(x)~g'(x)}{[g(x)]^2}$
