TITLE: Is there a differentiable function $f: {\mathbb R}_+ \to {\mathbb R}_+$ such that $f'(x) > f(x)^2$ for all $x$?
QUESTION [8 upvotes]: Is there a differentiable function $f: {\mathbb R}_+ \to {\mathbb R}_+$ such that $f'(x) > f(x)^2$ for all $x$?

It seems like the answer is no because such a function should have a vertical asymptote at some point.

REPLY [6 votes]: For any $x$ with $f(x) \neq 0$, your condition is the same as 
$${d \over dx} \bigg({1 \over f(x)}\bigg) < - 1 \tag 1$$
Note that $f'(x) > f(x)^2$ implies that $f(x)$ is strictly increasing. So there is at most one $x_0$ for which $f(x_0) = 0$. 
By the mean value theorem, for $x > x_0$ and $a > 0$ one has ${1 \over f(x + a)} - {1 \over f(x)} < -a$. Hence for $a$ large enough,  ${1 \over f(x + a)} < 0$, contradicting that the range of $f$ is contained in ${\mathbb R}^+$. Hence no such $f(x)$ can exist.