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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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Do you really mean "$x!$" ? – Arturo Magidin Jun 10 '12 at 3:07
$\,x!=\Gamma(x+1)\,\,,\,\,x+1\neq\,$ non-positive integer? – DonAntonio Jun 10 '12 at 3:09
@Riddler Yes, it is, but then I ask just what, apparently, Arturo meant to ask: how would you define $\,x!\,$ for a positive non-integer? – DonAntonio Jun 10 '12 at 3:17
Well, then is what I asked before! You're going to need the derivative of the gamma function, the polygamma: – DonAntonio Jun 10 '12 at 3:30
@Riddler: You can't restrict $x$ to positive integers and ask about derivatives. The computation of the derivative requires you to evaluate the function arbtirarily close to the point you want. Even if the "point you want" is an integer, points arbitrarily close to it will not be. So, no matter what, you'll end up having to deal with the derivative of the Gamma function. – Arturo Magidin Jun 10 '12 at 3:35
up vote 4 down vote accepted

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.

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+1 for mathematical nightmares! – AvatarOfChronos Jun 12 '12 at 14:50

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