Find a function such that $f^{-1}=f'$ Let $f:\Bbb{R}^+\rightarrow\Bbb{R}^+$ be a differentiable bijection and let $f$ satisfy: $f'=f^{-1}$ (where $f^{-1}$ denotes the inverse of $f$). Find $f$.
This comes from a facebook page "Mathematical theorems you had no idea existed, cause they're false". The negation of this statemed is given here, that is: There is no such function that satisfies $f'=f^{-1}$, but in the comments, there is a counterexample given:
$$g(x)=\varphi^{1-\varphi}x^{\varphi}$$
where $\varphi=\frac{1+\sqrt{5}}{2}$ is the golden ratio.
It's straightforward to check that $g'=g^{-1}$ holds. The OP claims that this solution is unique. Can someone come up with a way to derive the function $g$ or more functions satisfying this property? Also IS this really the unique solution?
 A: Here is an idea. Something along those lines should work.
Suppose we are trying to find out what the solutions looks like locally. Formally then, we expand in Taylor series
\begin{equation}
  f(x)=\sum_{n\geq 0}c_n x^n\implies f^\prime(x)=\sum_{n\geq 1}nc_n x^{n-1}.
\end{equation}
The condition we need satisfied is
\begin{equation}
f(f^\prime)(x)=x,
\end{equation}
or in other words
\begin{equation}
   \begin{aligned}
       c_0 &+ c_1\left(c_1+2c_2 x+3c_3 x^2+4c_4 x^4+\dots\right)\\
           &+ c_2\left(c_1+2c_2 x+3c_3 x^2+4c_4 x^4+\dots\right)^2\\
           &+ c_3\left(c_1+2c_2 x+3c_3 x^2+4c_4 x^4+\dots\right)^3\\
           &+ c_4\left(c_1+2c_2 x+3c_3 x^2+4c_4 x^4+\dots\right)^4\\
           &+ c_5 \left(c_1+2c_2 x+3c_3 x^2+4c_4 x^4+\dots\right)^5\\
           & \dots\\
           = &x
   \end{aligned}
\end{equation}
Let, for simplicity, the vector $C=(c_0,c_1,c_2,c_3,c_4,\dots)$. Then, the above relation is equivalent to
\begin{equation}
  \begin{aligned}
      &   C^T\cdot(1,c_1,c_1^2,c_1^3,\dots)=0, \\
      & C^T\cdot (0,2c_2,0,0,\dots)=2c_1c_2=1,\\
      & C^T\cdot (0,3c_3,(2c_2)^2,0,0,\dots)=0,\\
      & C^T \cdot (0,4c_4,(3c_3)^2,(2c_2)^3,0,\dots,)=0,\\
      & \dots.
  \end{aligned}
\end{equation}
We begine from the second equation above. We note that if we choose $c_1,c_2$, in the second equation above, then the rest $c_3,c_4,c_5$ are chosen inductively.
However, there is only one choice of $c_1,c_2$ such that the first equation is satisfied. Now since you have already found a solution, this solution is unique.
A: Too strange and incomprehensible for a simple comment, so I'll write it down as an answer.
You are looking for a function that satisfies the $f'=f^{-1}$ condition. What if we solve it like a differential equation? Then two solutions can be obtained:
$f_1=-\sqrt{2} \sqrt{c_1+x}$ and $f_2=\sqrt{2}\sqrt{c_1+x}$
At least Mathematica doesn't report anything about other solutions.
DSolve[f'[x] == f[x]^-1, f[x], x]

About the counterexample $g(x)=\varphi ^{1-\varphi } x^{\varphi }$, where $\varphi=\frac{1+\sqrt{5}}{2}$:
\[CurlyPhi] = (1 + Sqrt[5])/2;

g = Power[\[CurlyPhi], 1 - \[CurlyPhi]] Power[x, \[CurlyPhi]];

D[g, x] // N
    
Power[g, -1] // N

I saw that $g'≠g^{-1}$.
Plot[{Evaluate[D[g, x]], Evaluate[Power[g, -1]]}, {x, -5, 5}]


Don't take my post as an answer if you don't think it's correct. Just to try and understand your question better, I've done these calculations. And for the comment format, they did not fit.
