Is there a differentiable function $f: {\mathbb R}_+ \to {\mathbb R}_+$ such that $f'(x) > f(x)^2$ for all $x$? 
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.
 A: 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. 
A: No. Any such function has a vertical asymptote. The solution to $g'(x)=g(x)^2$ is $g(x)=\frac{-1}{x+c}$. In particular, suppose we had an $f$ satisfying $f'(x)>f(x)^2$ for all $x$. Then, choose $g$ to satisfy:
$$g(1)=f(1)$$
$$g'(x)=g(x)^2.$$
It is clear that $g$, being of the before-mentioned form, will have an asymptote. In fact, to be explicit, we have
$$g(x)=\frac{-1}{x-1-\frac{1}{f(1)}}$$
which has an asymptote at $1+\frac{1}{f(1)}$. However, it is clear that $f(x)>g(x)$ for all $x>1$. Thus, $f$ must have an asymptote at or before $1+\frac{1}{f(1)}$.
