Find $f(x)$ such that $f^\prime(x) = f(f(x))$

Question

Is there such a differentiable function $f: R\rightarrow R$ that for

each real $x$ we have $f (x)> 0$ and $f' (x) = f (f (x))$;

Answer 1

No such function does exist.

Assume the contrary. Since $f'(x)=f(f(x))>0$ for any $x\in\Bbb R$, we know that $f$ is increasing.

Mean Value Theorem implies that there exists some $c\in(-1,0)$ such that $$f(f(c))=f'(c)=f(0)-f(-1)<f(0)$$ Since $f$ is increasing, $f(c)<0$, a contradiction.