Is there such a differentiable function $f: R\rightarrow R$ that for
each real $x$ we have $f (x)> 0$ and $f' (x) = f (f (x))$;
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I don't see why such a function shouldn't exist, but it might not be expressible in terms of elementary functions. You might want to read about functional square roots. (There's also a lot of questions on stackexchange about it). – Henrik supports the community Mar 11 '17 at 09:21
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A solution $f$ of the functional equation $f(f(x))=g(x)$ is called an iterative root of $g$. – mlc Mar 11 '17 at 09:46
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I don't really see the point of voting down this question, even though this question is short – edm Mar 11 '17 at 10:35
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I agree with @edm: hence, my upvote. – mlc Mar 11 '17 at 18:55
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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.
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