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?
 A: 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 $\phi^2=\phi+1$:
$$
\begin{eqnarray}
f''(x)&=&\frac1{f'(f'(x))}\\
\phi^{1-\phi}x^{\phi-2}&=&\left(\phi^{2-\phi}\left(\phi^{2-\phi} 
x^{\phi-1}\right)^{\phi-1}\right)^{-1}\\
&=&\left(\phi^{2-\phi}\left(\phi^{3\phi-2-\phi^2}x^{\phi^2-2\phi+1}\right)\right)^{-1}\\
&=&\left(\phi^{2\phi-\phi-1}x^{\phi+1-2\phi+1}\right)^{-1}\\
&=&\left(\phi^{\phi-1}x^{-\phi+2}\right)^{-1}\\
\end{eqnarray}
$$
A: 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.
A: When you set $f^{-1}(x)=f'(x)$ these functions have the property that
$$
        f^{-1}(x)=\int\frac{1}{f'(f^{-1}(x))}\,dx + c. \tag{1}
$$
from "Inverse functions and differentiation".
A: Start with $\dot{y}=y^{-1}$  Apply Lie Group $G(x,y)=(\lambda x,\lambda^\beta y)\lambda_o=1$
Stabilizers for this group include $\mu=\frac{y}{x^\frac{1}{2}}$ and $\nu=x^{\frac{1}{2}}\dot{y}$.  Multiplying the DEQ by $x^{\frac{1}{2}}$ we have $x^{\frac{1}{2}}\dot{y}=\bigg(\frac{x^{\frac{1}{2}}}{y}\bigg)$ which makes our DEQ, in stabilizer form, $\nu=\mu^{-1}$.  Since $$x\frac{d\mu}{dx}=\nu -\beta\mu = \mu^{-1} -\frac{1}{2}\mu$$ $$\frac{d\mu}{\mu^{-1}-\frac{1}{2}\mu}=\frac{dx}{x}$$ Integrating this gives $$-ln(\mu^2-2)=lnx+lnC$$ Substituting for $\mu$ and rearranging arrives at the solution: $$y=\sqrt{2x+C}$$
