Answer
$h'(x) = \frac{[f(x)]^2g'(x)+f'(x)[g(x)]^2}{[f(x)+g(x)]^2}$
Work Step by Step
$h(x) = \frac{f(x)g(x)}{f(x)+g(x)}$
$h'(x) = \frac{[f'(x)g(x)+f(x)g'(x)]\cdot [f(x)+g(x)]-f(x)g(x)\cdot [f'(x)+g'(x)]}{[f(x)+g(x)]^2}$
$h'(x) = \frac{[f(x)f'(x)g(x)+f(x)f(x)g'(x)+g(x)f'(x)g(x)+g(x)f(x)g'(x)]-[f(x)g(x)f'(x) +f(x)g(x)g'(x)]}{[f(x)+g(x)]^2}$
$h'(x) = \frac{[f(x)f(x)g'(x)+g(x)f'(x)g(x)]}{[f(x)+g(x)]^2}$
$h'(x) = \frac{[f(x)]^2g'(x)+f'(x)[g(x)]^2}{[f(x)+g(x)]^2}$
