Answer
$f'(x) =\dfrac{4}{(1-x)^{2}}$ and $ f'(-2) = \dfrac{4}{9}$
Work Step by Step
Consider $f(x) = \dfrac{x+3}{1-x} \implies f(-2) = \dfrac{1}{3}$
$f'(x) = \lim\limits_{h \to 0}\dfrac{\dfrac{(x+h)+3}{1-(x+h)}-\dfrac{x+3}{1-x}}{h}= \lim\limits_{h \to 0}\dfrac{4h}{h(1-x)(1-x-h)}=\dfrac{4}{(1-x)(1-x-0)}=\dfrac{4}{(1-x)^{2}}$
$ f'(-2) =\dfrac{4}{(1-(-2))^{2}}= \dfrac{4}{9}$
Hence, $f'(x) =\dfrac{4}{(1-x)^{2}}$ and $ f'(-2) = \dfrac{4}{9}$
