Answer
$\dfrac{-1}{(x-1)^{2}}$
Work Step by Step
Consider $g(x) = \dfrac{x}{x-1}$
We will apply definition of derivative, such as:
$g'(x) = \lim\limits_{z \to x}\dfrac{\dfrac{z}{z-1} - \dfrac{x}{x-1}}{(z-x)} = \lim\limits_{z \to x}\dfrac{(x-z)}{(z-x)(x-1)(z-1)}$
$\implies \lim\limits_{z \to x}\dfrac{-1}{(x-1)(z-1)}=-\dfrac{1}{(x-1)(x-1)}$
Hence, $g'(x) = \dfrac{-1}{(x-1)^{2}}$
