Answer
$\frac{1}{(2-x)^{2}}$
Work Step by Step
$\frac{dy}{dx}=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\frac{\frac{1}{2-(x+h)}-\frac{1}{2-x}}{h}$
$=\lim\limits_{h \to 0}=\frac{\frac{(2-x)-[2-(x+h)]}{[2-(x+h)](2-x)}}{h}$
$=\lim\limits_{h \to 0}\frac{h}{[2-(x+h)](2-x)h}$
$\lim\limits_{h \to 0}\frac{1}{[2-(x+h)](2-x)}$
$=\lim\limits_{h \to 0}\frac{1}{(2-x)(2-x)}$
$=\frac{1}{(2-x)^{2}}$
