I have a theory that uses the gamma function:
$$\Gamma(n)=\int_0^\infty x^{n-1}e^{-x} \space dx$$
Then I was inclined to think that perhaps the derivative is:
$$x^{n-1}e^{-x}$$
But I'm not sure we can just drop the integral along with the bounds to get the derivative. Then I thought about taking the limit:
$$\lim_{x\to\infty}x^{n-1}e^{-x}$$
But now we can't specify at what $x$ value we want to get the rate of change of. At this point I feel like I can't get any further on my own and would appreciate some insight.
EDIT: Looking for derivative in terms of $n$ actually.
Here is how to calculate it: you have to move the derivative into the integral: \begin{align} \frac{d}{dn}\Gamma(n) &=\frac{d}{dn}\int_0^\infty x^{n-1}e^{-x}\,dx\\ &=\int_0^\infty \frac{d}{dn}x^{n-1}e^{-x}\,dx\\ &=\int_0^\infty e^{-x}\frac{d}{dn}e^{(n-1)\ln(x)}\,dx\\ &=\int_0^\infty e^{-x}\cdot e^{(n-1)\ln(x)}\ln(x)\,dx\\ &=\int_0^\infty x^{n-1}e^{-x}\ln(x)\,dx\\ \end{align} and so we have $$\Gamma'(n)=\int_0^\infty x^{n-1}e^{-x}\ln(x)\,dx$$