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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?

Han de Bruijn
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    Yes, that's correct. You can easily check that with Mathematica or WolframAlpha – polfosol Apr 24 '21 at 13:37
  • The number $N=\sqrt[3]{-1}$ where the answers are the complex roots. – Тyma Gaidash Apr 24 '21 at 13:40
  • @polfosol: Thanks for the response. Usually I check my mathematics with MAPLE. But there is only an old version (8) of it available on my PC, which doesn't perform very well on complex numbers. – Han de Bruijn Apr 24 '21 at 14:47
  • WolframAlpha is technically free for such calculations. And in case you want to dive a little deeper into the world of Mathematica, there is the online version, also free ;) – polfosol Apr 24 '21 at 14:59

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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: $$ y = -i.e^{\pi/\sqrt{3}/2}.x^{1/2+i.\sqrt{3}/2} \\ y = +i.e^{\pi/\sqrt{3}/2}.x^{1/2-i.\sqrt{3}/2} $$

Han de Bruijn
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