1. ## Inverse and derivative

Give an example of a differentiable function $f: (0, \infty) \longrightarrow \mathbb{R}$ such that $f '(x)=f^{-1}(x), \ \forall x > 0.$ Can you prove that there are no other examples?

2. Originally Posted by NonCommAlg
Give an example of a differentiable function $f: (0, \infty) \longrightarrow \mathbb{R}$ such that $f '(x)=f^{-1}(x), \ \forall x > 0.$ Can you prove that there are no other examples?
Here is one example:

Let $f(x)=ax^n$, then
$f'(x)=anx^{n-1}$
$f(f'(x))=f(anx^{n-1})=a(anx^{n-1})^n=a^{n+1}n^nx^{n(n-1)}=x$

Therefore
$n(n-1)=1, a^{n+1}n^n=1$
$n_1=\frac{1+\sqrt{5}}{2}$, $a_1=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}}$
$n_2=\frac{1-\sqrt{5}}{2}$, $a_2=\left(\frac{1-\sqrt{5}}{2}\right)^{\frac{1+\sqrt{5}}{2}}$ (This one is not real).

So,
$f(x)=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}}x^{\frac{1+\sqrt{5}}{2}}$

3. Originally Posted by luobo
Here is one example:

Let $f(x)=ax^n$, then
$f'(x)=anx^{n-1}$
$f(f'(x))=f(anx^{n-1})=a(anx^{n-1})^n=a^{n+1}n^nx^{n(n-1)}=x$

Therefore
$n(n-1)=1, a^{n+1}n^n=1$
$n_1=\frac{1+\sqrt{5}}{2}$, $a_1=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}}$
$n_2=\frac{1-\sqrt{5}}{2}$, $a_2=\left(\frac{1-\sqrt{5}}{2}\right)^{\frac{1+\sqrt{5}}{2}}$ (This one is not real).

So,
$f(x)=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}}x^{\frac{1+\sqrt{5}}{2}}$

Is it the unique function for this ?

4. Originally Posted by simplependulum
Is it the unique function for this ?
yes! luobo's example, i.e. $f(x)=\varphi^{1 - \varphi}x^{\varphi},$ where $\varphi=\frac{1+\sqrt{5}}{2},$ is the only function that satisfies the condition. now the question is: why?