Does $f'(x) \in o(g'(x))$ imply $f(x) \in o(g(x))$ for monotonically increasing $f$ and $g$?

Question

The title says it all. This seems intuitively true to me, but I'm not sure how one would go about proving this.

(I'm asking because I'm trying to show that $x^n \in o(x^{n+1})$ for all natural $n$, and I was wondering if you could just take derivatives and use induction.)

Answer 1

It will be a lot easier to divide $x^n$ by $x^{n+1}$, see what happens as $x$ gets big.

Anyway, the result about derivatives is not correct. Let $f(x)=1-\frac{1}{x^2}$ and let $g(x)=1-\frac{1}{x}$. Then $f'(x)\in o(g'(x))$ but $f(x)\not\in o(g(x))$.