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Could someone explain how I would find the derivative of the following function? I am completely lost: $f(x) = e^{i(x!^{\log x})}$

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    @ArturoMagidin I guess I don't really have an choice then. If you are aware of one, could you please provide a solution with the derivative of the Gamma function?2012-06-10

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For complex $\,z\,$ with $\,\operatorname{Re}(z)>0\,$ , $\Gamma'(z)=\int_0^\infty t^{z-1}e^{-t}\log t\,dt$ so $\left(e^{ix!^{\log x}}\right)'=e^{ix!^{\log x}}\left(ix!^{\log x}\right)'$ Now, since $\,\displaystyle{x!^{\log x}=e^{\log x\log(x!)}}\,\,and\,\,\log x!=\log\Gamma(x+1)$ , we get$\left(ie^{\log x\log x!}\right)'=i\left[\frac{\log x!}{x}+\log x\frac{\Gamma'(x+1)}{\Gamma(x+1)}\right]e^{\log x\log x!} $Now put together all the above and have sweet mathematical nightmares.