TITLE: Function whose inverse is also its derivative?
QUESTION [17 upvotes]: What are some good examples of a function  $f : \mathbb{R} \to \mathbb{R}$ where its derivative is equal to its inverse? I attempted to find a monomial that satisfied it by starting with $f(x) = ax^b$ and showing that $f^{-1}(x) = f'(x) \implies b-1=\frac{1}{b} \implies b=\phi$ and got
$$f(x) = \frac{x^\phi}{\sqrt[\phi]{\phi}}$$
Which seems to work according to WolframAlpha, but I'm having trouble double-checking it. Any other ideas?

REPLY [3 votes]: What you have given is only a function from $(0,\infty)$ to $(0,\infty)$, not $\mathbb R$ to $\mathbb R$.  See Function satisfying $f^{-1} =f'$ on MathOverflow for some other ideas and more thorough analysis with this restriction to positive reals.
For the domain $\mathbb R$, no solution exists.  A continuous injective $f:\mathbb R\to\mathbb R$ must be monotone, which implies that its derivative cannot change sign, but $f^{-1}$ would include both positive and negative numbers in its range.

Here's what it looks like in Desmos if you extend your answer following Mario Carneiro's advice in a comment (which I didn't understand at first):

Piecing these functions together gives an invertible map from $\mathbb R$ onto $\mathbb R$ such that $f'(x) = f^{-1}(x)$  when $f'(x)$  exists, and $f'(0)$ doesn't exist, but the right-hand derivative $\lim\limits_{h\to 0+}\dfrac{f(h)-f(0)}{h}$ exists and equals $0=f^{-1}(0)$.   Considering that a differentiable solution is impossible, this is pretty good.