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

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

so $f'(x) = \frac{-1 ~(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}$

as (1 - x)² > 0 always ⇒ f(x) > 0 always

so for x > 1 → f(x) is increasing