Answer
See the explanation.
Work Step by Step
Let $f(x)=\frac{1}{x}$.
Using the definition of a derivative,
$f'(x)=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$ (Simplify the numerator)
$=\lim\limits_{h \to 0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}$
$=\lim\limits_{h \to 0}\frac{\frac{-h}{x(x+h)}}{h}$ (Cancel $h$)
$=\lim\limits_{h \to 0}\frac{-1}{x(x+h)}$ (Evaluate the limit by direct substitution)
$=\frac{-1}{x(x+0)}$
$=\frac{-1}{x^2}$
So, $f'(x)=-\frac{1}{x^2}$.
