Question

What's the derivative of the factorial function?

y=x!

Answer 1

My suspicion is that the derivative of the factorial function is itself. However I'm not really that sure about the whole gamma function nonsense. Here's my rationale:

\[n!=\int\limits_{0}^{\infty} x^n e^{-x}dx\] \[\frac{ d }{ dn } (n!)=\frac{ d }{ dn } ( \int\limits_{0}^{\infty} x^n e^{-x}dx)=n \int\limits_{0}^{\infty} x^{n-1} e^{-x}dx=n*(n-1)!=n!\]

But then things change, since if you actually plug in a value it seems like there could be a problem with the n-1 in the integral.

Answer 2

I think things get a little more complicated than that.\[\large \frac{ d }{ dn } (n!)=\frac{ d }{ dn } \int\limits\limits_{0}^{\infty} x^n e^{-x}dx= \int\limits\limits_{0}^{\infty} \color{red}{x^{n}} e^{-x}\color{red}{\ln x}\;dx\]I can't really make sense of all this. But I think that's what our derivative is giving us, yes?

Answer 3

Yes, you're completely right haha. What a rookie mistake on my part. >_>

Answer 4

Maybe it's easier if we look at a more general integral version.

\[\frac{ n! }{ t^{n+1} }=\int\limits_{0}^{\infty} x^n e^{-tx}dx\] and then take the derivative of that or something.

Or actually if we use integration by parts can we solve the original one?

Answer 5

I'm just sorta having fun here, there's actually a really quick easy way to derive this formula if you're interested.

Answer 6

But I don't know how to find the derivative of it though, that's another problem altogether.