Is there any function f which is only differentiable on $(0,\infty)$ and such that $f^{-1} = f'$?

Question

Is there any function f which is only differentiable on $(0,\infty)$ and such that $f^{-1} = f'$?

I thinks there exists no such function. I thought about constant, exponential, trigonometric function, etc., but I didn't find any function which is only differentiable on $(0,\infty)$ and such that $f^{-1} = f'$.

Is it true?

Answer 1

There is at least one such function. Let us start by assuming that $f(x)=kx^\alpha$, for some $k,\alpha>0$. Then $f'(x)=k\alpha x^{\alpha-1}$ and $f^{-1}(x)=k^{-\frac1\alpha}x^\frac1\alpha$. So, when do you have$$\bigl(\forall x\in(0,+\infty)\bigr):k\alpha x^{\alpha-1}=k^{-\frac1\alpha}x^\frac1\alpha?$$We must have $\alpha-1=\frac1\alpha$, which means that $\alpha=\varphi$ (the golden ratio). And then $k$ must be such that $k^{-\frac1\varphi}=k\varphi$, wich is equivalent to $k^{-\frac1\varphi-1}=\varphi$. Numerically, $k\approx0.7427$.

If you allow $\alpha$ to be negative, the same approach leads to another solution, of the form $x\mapsto kx^{-\frac1\varphi}.$