Answer
$f′(x)=\frac{-2}{x^{3}}$
Work Step by Step
$f(x)=7$
$f′(x)=\lim\limits_{h \to 0}\frac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}$
$f′(x)=\lim\limits_{h \to 0}\frac{x^{2}-(x+h)^{2}}{(x)^{2}(x+h)^{2}(h)}$
$f′(x)=\lim\limits_{h \to 0}\frac{x^{2}-x^{2}-2xh+h^{2}}{(x)^{2}(x+h)^{2}(h)}$
$f′(x)=\lim\limits_{h \to 0}\frac{h(-2x+h)}{(x)^{2}(x+h)^{2}(h)}$
$f′(x)=\lim\limits_{h \to 0}\frac{-2x+h}{(x)^{2}(x+h)^{2}}$
$f′(x)=\frac{-2x+0}{x^{2}(x+0)^{2}}$
$f′(x)=\frac{-2}{x^{3}}$
