Answer
\[\frac{dy}{dx}=\frac{f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right)}{2\sqrt{f\left( x \right)g\left( x \right)}}\]
Work Step by Step
\[\begin{align}
& y=\sqrt{f\left( x \right)g\left( x \right)} \\
& \text{Rewrite} \\
& y={{\left[ f\left( x \right)g\left( x \right) \right]}^{1/2}} \\
& \text{Differentiate both sides with respect to }x \\
& \frac{dy}{dx}=\frac{d}{dx}\left[ {{\left( f\left( x \right)g\left( x \right) \right)}^{1/2}} \right] \\
& \frac{dy}{dx}=\frac{1}{2}{{\left[ f\left( x \right)g\left( x \right) \right]}^{-1/2}}\frac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right] \\
& \text{Use the product rule} \\
& \frac{dy}{dx}=\frac{1}{2}{{\left[ f\left( x \right)g\left( x \right) \right]}^{-1/2}}\left( f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right) \right) \\
& \text{Simplifying} \\
& \frac{dy}{dx}=\frac{f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right)}{2\sqrt{f\left( x \right)g\left( x \right)}} \\
\end{align}\]
