Answer
\[\frac{f'\left( x \right)g\left( x \right)-f\left( x \right)\text{ }g'\left( x \right)\text{ }}{g{{\left( x \right)}^{2}}}\]
Work Step by Step
\[\begin{align}
& \frac{d}{dx}\left( \frac{f\left( x \right)}{g\left( x \right)} \right) \\
& \text{We can write }\frac{f\left( x \right)}{g\left( x \right)}\text{ as }f\left( x \right)g{{\left( x \right)}^{-1}},\text{ then} \\
& \frac{d}{dx}\left( \frac{f\left( x \right)}{g\left( x \right)} \right)=\frac{d}{dx}\left( \text{ }f\left( x \right)g{{\left( x \right)}^{-1}} \right) \\
& \text{By the product rule} \\
& \text{ }=f\left( x \right)\frac{d}{dx}\left( \text{ }g{{\left( x \right)}^{-1}} \right)+\text{ }g{{\left( x \right)}^{-1}}\frac{d}{dx}\left( f\left( x \right) \right) \\
& \text{ }=-f\left( x \right)\text{ }g{{\left( x \right)}^{-2}}\frac{d}{dx}\left( g\left( x \right) \right)+\text{ }g{{\left( x \right)}^{-1}}f'\left( x \right) \\
& \text{ }=-f\left( x \right)\text{ }g{{\left( x \right)}^{-2}}g'\left( x \right)+\text{ }g{{\left( x \right)}^{-1}}f'\left( x \right) \\
& \text{Simplifying} \\
& \text{ }=\frac{-f\left( x \right)\text{ }g'\left( x \right)}{g{{\left( x \right)}^{2}}}+\frac{\text{ }f'\left( x \right)}{g\left( x \right)} \\
& \text{ }=\frac{-f\left( x \right)\text{ }g'\left( x \right)+\text{ }f'\left( x \right)g\left( x \right)}{g{{\left( x \right)}^{2}}} \\
& \text{ }=\frac{f'\left( x \right)g\left( x \right)-f\left( x \right)\text{ }g'\left( x \right)\text{ }}{g{{\left( x \right)}^{2}}} \\
\end{align}\]
