Question Number 1724 by Hasan Mohamed last updated on 03/Sep/15
$$ \\ $$
Answered by 123456 last updated on 04/Sep/15
$${f}\left({x}\right)={x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$${f}'\left({x}\right)={x}!\psi\left({x}+\mathrm{1}\right) \\ $$$$\psi\left({x}\right)=\frac{{d}}{{dx}}\mathrm{ln}\:\Gamma\left({x}\right) \\ $$