ODE with inverse function. Solve $f^{-1}(x)=f'(x)$. 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 $x&lt;0$ 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.
Partial answer, in order to confirm the Omnomnomnom's assumption that some solutions on the form $f(x)=ax^b$ do exist.

$$f^{-1}(x)=f'(x)\qquad \text{on }x&gt;0$$
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 : $\quad f(x)=\left(\frac{2\:x}{1-\sqrt{5}}\right)^{\frac{1-\sqrt{5}}{2}}$
Make the substitution $y = f(x)$ to rewrite this equation as
$$
y = f'(f(y))
$$
Rewrite this as
$$
y\,f'(y) = f'(f(y)) \,f'(y)
$$
Now, integrate both sides with respect to $y$.  Using integration by parts, note that
$$
\int y\,f'(y) \,dy = y\,f(y) - \int f'(y)\,dy = (y - 1)f(y) + C
$$
Using substitution, note that
$$
\int f'(f(y)) \,f'(y)\,dy = 
\int f'(u)\,du = f(u) + C = f(f(y)) + C
$$
Conclude that any $f$ satisfying our original equation must satisfy
$$
(y - 1)f(y) = f(f(y)) + C
$$
In terms of $x$, this says
$$
(f^{-1}(x) - 1)x = f(x) + C
$$
This isn't quite a solution, but at least we don't have any derivatives here.



Things to be said from here:

Note that with $y = 1$, we have
$$
0 \cdot f(1) = f(f(1)) + C \implies C = -f(f(1))
$$
Note also that if $f(y) = 0$ for any $y$, that $y$ must satisfy
$$
f(f(y)) = f(f(1))
$$ 
but if $f$ is "invertible" and therefore $1$ to $1$, then this should never happen.