If f(x)=1/(1-x), then for x>1, f(x) is:

$f(x)=\frac{1}{1-x}=(1-x)^{-1}$

so, $f'(x) = (-1)×(1-x)^{-2}×\frac{d}{dx}(1-x)$

$=(-1)×(-1)×(1-x)^{-2}$

$=\frac{1}{(1-x)^2}>0$ always

$f(x)$ is increasing always