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