Answer
$$\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right] = \frac{f^{\prime}(x) g(x)-f(x) g^{\prime}(x)}{[g(x)]^{2}}$$
Work Step by Step
We evaluate the derivative as follows:
\begin{aligned}
\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right] &=\frac{1}{h}\lim _{h \rightarrow 0} \frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)} \\
&=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{g(x+h) g(x) h} \\
&=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{g(x+h) g(x) h} \\
&=\lim _{h \rightarrow 0} \frac{[(x+h)-f(x)] g(x)-f(x)[(g(x+h)-g(x)]}{g(x+h) g(x) h} \\
&=\lim _{h \rightarrow 0} \frac{\frac{f(x+h)-f(x)}{h} g(x)-f(x) \frac{g(x+h)-g(x)}{h}}{g(x+h) g(x)}\\
&=\frac{ \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} g(x)-f(x) \lim _{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}}{ \lim _{h \rightarrow 0} g(x+h) g(x)}\\
&= \frac{f^{\prime}(x) g(x)-f(x) g^{\prime}(x)}{[g(x)]^{2}}
\end{aligned}
