The question Derivative and inverse
is accompanied with an interesting
answer
by the same author and a nice video called
Inverse Prime equals Prime Inverse.
The video ends with the formula $C = N^{N(N-2)}$, which is simplified further in the author's answer to the question.
The problem is that I have not been able to reproduce this simplification. This is what I have done.
First solution:
$$
N = (1+i\sqrt{3})/2=e^{i\pi/3} \\
N(N-2)=(N^2-N+1)-(N+1)=-(N+1) \\
C = N^{-(N+1)} = \left(e^{i\pi/3}\right)^{(-3/2-i\sqrt{3}/2)} = e^{-i\pi/2}e^{\pi/(2\sqrt{3})}=-i\,e^{\pi/(2\sqrt{3})}
$$
Second solution:
$$
N = (1-i\sqrt{3})/2=e^{-i\pi/3} \\
C = N^{-(N+1)} = \left(e^{-i\pi/3}\right)^{(-3/2+i\sqrt{3}/2)} = e^{i\pi/2}e^{\pi/(2\sqrt{3})}=+i\,e^{\pi/(2\sqrt{3})}
$$
Is this correct?
Is there a function who's derivative of the inverse equals the inverse of its derivative?
$$
\frac{d}{dx}f^{\langle-1\rangle}(x)=\left(\frac{d}{dx}f(x)\right)^{\langle-1\rangle}
$$
With functions of the form $y=Cx^N$ the answer is affirmative for complex $N$ and $C$.
The two solutions (when simplified correctly) are:
$$
y = -i.e^{\pi/\sqrt{3}/2}.x^{1/2+i.\sqrt{3}/2} \\
y = +i.e^{\pi/\sqrt{3}/2}.x^{1/2-i.\sqrt{3}/2}
$$