Answer
$$h'(x)=\frac{f^{\prime}(x)[g(x)]^{2}+g^{\prime}(x)[f(x)]^{2}}{[f(x)+g(x)]^{2}} $$
Work Step by Step
Given
$$ h(x)=\frac{f(x)g(x)}{f(x)+g(x)}$$
Then
$\begin{align*}
h^{\prime}(x) &=\frac{[f(x)+g(x)]\left[f(x) g^{\prime}(x)+g(x) f^{\prime}(x)\right]-f(x) g(x)\left[f^{\prime}(x)+g^{\prime}(x)\right]}{[f(x)+g(x)]^{2}} \\ &=\frac{[f(x)]^{2} g^{\prime}(x)+f(x) g(x) f^{\prime}(x)+f(x) g(x) g^{\prime}(x)+[g(x)]^{2} f^{\prime}(x)}{[f(x)+g(x)]^{2}}-\frac{f(x) g(x) f^{\prime}(x)+f(x) g(x) g^{\prime}(x)}{[f(x)+g(x)]^{2}} \\
&=\frac{f^{\prime}(x)[g(x)]^{2}+g^{\prime}(x)[f(x)]^{2}}{[f(x)+g(x)]^{2}} \end{align*}$
