Answer
$g'(x) = \frac{-1}{(x-1)^{2}}$
Work Step by Step
$g(x) = \frac{x}{x-1}$
$g'(x) = \lim\limits_{z \to x}\frac{\frac{z}{z-1} - \frac{x}{x-1}}{z-x}$
$g'(x) = \lim\limits_{z \to x}\frac{1}{z-x}\frac{x-z}{(x-1)(z-1)}$
$g'(x) = \lim\limits_{z \to x}\frac{-1}{(x-1)(z-1)}$
$g'(x) = \frac{-1}{(x-1)^{2}}$
