Function whose inverse is also its derivative?

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?

Solution 1:

Here is an alternative: You can use $$ f^{-1}(x)=\int\frac{1}{f'(f^{-1}(x))}\,dx + c. \tag{1} $$ from "Inverse functions and differentiation".

Set $f^{-1}(x)=f'(x)$ and for simplicity take the derivative of $(1)$. You get $$ f''(x)=\frac1{f'(f'(x))}. $$ Now put in $f(x)=\phi^{-\frac1\phi}x^\phi$. Using $\phi-1=\frac1\phi$, you'll get $$ f'(x)=\phi^{2-\phi}x^{\phi-1}\\ f''(x)=\phi^{1-\phi}x^{\phi-2} $$ Now put it all together and use $\phi^2=\phi+1$: $$ \begin{eqnarray} f''(x)&=&\frac1{f'(f'(x))}\\ \phi^{1-\phi}x^{\phi-2}&=&\left(\phi^{2-\phi}\left(\phi^{2-\phi} x^{\phi-1}\right)^{\phi-1}\right)^{-1}\\ &=&\left(\phi^{2-\phi}\left(\phi^{3\phi-2-\phi^2}x^{\phi^2-2\phi+1}\right)\right)^{-1}\\ &=&\left(\phi^{2\phi-\phi-1}x^{\phi+1-2\phi+1}\right)^{-1}\\ &=&\left(\phi^{\phi-1}x^{-\phi+2}\right)^{-1}\\ \end{eqnarray} $$

Solution 2:

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.