Answer
$f'(x)=-\frac{4}{x^3}$
The domain of $f'$: $(-\infty,0)\cup(0,\infty)$.
Work Step by Step
Use the definition of a derivative.
$f'(x)=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\frac{\frac{2}{(x+h)^2}-\frac{2}{x^2}}{h}$
$=\lim\limits_{h \to 0}\frac{\frac{2x^2-2(x+h)^2}{(x+h)^2x^2}}{h}$
$=\lim\limits_{h \to 0}\frac{\frac{2x^2-2x^2-4xh-2h^2}{(x+h)^2x^2}}{h}$
$=\lim\limits_{h \to 0}\frac{\frac{-4xh-2h^2}{(x(x+h))^2}}{h}$
$=\lim\limits_{h \to 0}\frac{h(-4x-2h)}{h(x(x+h))^2}$
$=\lim\limits_{h \to 0}\frac{-4x-2h}{(x(x+h))^2}$
$=\frac{-4x-2\cdot 0}{(x(x+0))^2}$
$=\frac{-4x}{x^4}$
$=\frac{-4}{x^3}$
Thus, $f'(x)=-\frac{4}{x^3}$.
The domain of $f'$ is $(-\infty,0)\cup(0,\infty)$.
