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