TITLE: ODE with inverse function. Solve $f^{-1}(x)=f'(x)$.
QUESTION [5 upvotes]: Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that $$f^{-1}(x)=f'(x)$$
I think one such function would be of the form $f(x)=ax^b$. But then $b$ would be irrational and when $x\lt0$ this causes problems. So I guess letting $f(x)=-a(-x)^b$ for $x<0$ might work. But that's only one possibble solution, what are all the solutions?
Edit: With the comment of Joey Zou, the domain and range has been changed to $(0,\infty)$.
Edit: I already know the solution of the form $ax^b$. However, what I'm really asking is whether it is the unique solution.

REPLY [2 votes]: Partial answer, in order to confirm the Omnomnomnom's assumption that some solutions on the form $f(x)=ax^b$ do exist.
$$f^{-1}(x)=f'(x)\qquad \text{on }x>0$$
Search of particular solutions on the form $\quad f(x)=ax^b$ :
$f=ax^b \quad\to\quad x=\left(\frac{f}{a}\right)^{1/b}$
Hence the inverse fonction of $f(x)$ expressed as a function of $x$ is :
$$f^{-1}(x)=\left(\frac{x}{a}\right)^{1/b}$$
to be not confused with $x$ expressed as a function of $f$.
If a solution of the form $f(x)=ax^b$ exists, it must satisfy the equation :
$$f^{-1}(x)=f'(x)=\left(\frac{x}{a}\right)^{1/b}=abx^{b-1}$$
First condition : $x^{1/b}=x^{b-1}\quad\to\quad b^2-b-1=0$
$$b=\frac{1\pm\sqrt{5}}{2}$$
Second condition : $\left(\frac{1}{a}\right)^{1/b}=ab \quad\to\quad a=b^{-\frac{b+1}{b}}$
So, two different particular solutions are obtained :
First : $\quad f(x)=\left(\frac{2\:x}{1+\sqrt{5}}\right)^{\frac{1+\sqrt{5}}{2}}$
Second : $\quad f(x)=\left(\frac{2\:x}{1-\sqrt{5}}\right)^{\frac{1-\sqrt{5}}{2}}$