TITLE: Can there be an injective function whose derivative is equivalent to its inverse function?
QUESTION [59 upvotes]: Let's say $f:D\to R$ is an injective function on some domain where it is also differentiable. For a real function, i.e. $D\subset\mathbb R, R\subset\mathbb R$, is it possible that $f'(x)\equiv f^{-1}(x)$?
Intuitively speaking, I suspect that this is not possible, but I can't provide a reasonable proof since I know very little nothing about functional analysis. Can anyone provide a (counter)example or prove that such function does not exist?

REPLY [8 votes]: Since someone mentioned whether the answer given by Robert Israel / Greg Martin might be unique, I thought it is worth noting that the function
$$f(x)=-\frac{1}{\phi^\phi}(-x)^{-\frac{1}{\phi}},\quad x<0$$
where $\phi$ is the golden ratio, has the same property on $D=(-\infty,0)$, i.e. $f'(x)\equiv f^{-1}(x)$.
Edit- So if we define:
$$f:\mathbb R\to\mathbb R\\x\mapsto a(x/a)^a$$
where $a=\frac{1+\sqrt{5}\text{ sign}(x)}{2}$, we would have a bijection on $\mathbb R$ with that nice property (yay!...).