$f$ has a derivative for every $x \in (0,\infty)$
$f'(x) > x$ for every $x>0$
Can I prove that $f(x+1) - f(x)$ is a monotonic increasing function?
From Lagrange I know that in every $I = [x,x+1]$ where $x>0$, $f(x+1) - f(x) > x$.
$f$ has derivative so $f'(x+1) - f'(x)$ is defined for $x>0$.
But, How can I show that $f'(x+1) - f'(x) > 0$?
Edit: Thank you al for your answers! If I want to prove a weaker statement, that there exists some $M \in \mathbb R$ such that $f'(x+1) - f'(x)$ is increasing in $(M,\infty)$. Is it, then, can be proved?