Let us denote the function $f(x, n)$ (known as the falling factorial) as follows
$$f(x,n) = \prod_{k=0}^{n-1}(x-k)$$
Note that $$f(n,n) = n!$$
I would like to find the derivative of $f(x,n)$. My attempt written out in full is shown below:
$$\ln f(x,n) = \ln\prod_{k=0}^{n-1}(x-k)$$
$$\ln f(x,n) = \sum_{k=0}^{n-1}\ln(x-k)$$
$$\frac{f'(x,n)}{f(x,n)} = \sum_{k=0}^{n-1}\frac{1}{(x-k)}$$
and finally
$$f'(x,n) = \sum_{k=0}^{n-1}\frac{f(x,n)}{(x-k)}$$
If we let $x=n$ we have that
$$f'(n,n) = n!\sum_{k=0}^{n-1}\frac{1}{(n-k)}$$
This naïvely lead part of me to conclude that the 'elementary derivative' of $x!$ is given by $f'(n,n)$. However, I know of the gamma function and its derivative, namely that
$$\Gamma'(n+1) = n!\left(-\gamma + \sum_{k=0}^{n}\frac{1}{k}\right)$$
Comparing the two, they are almost mathematically identical except for the $\gamma$ constant. I am unsure of why this is. My calculation seems correct although it seems like it can't be. Where have I gone wrong?
Edit: I want to clarify that I by no means am saying that the derivatives of both $\Gamma$ and $f$ are equal, I am saying that I expect them to reduce to equal form when considering the 'elementary derivative' of $x!$ and am wondering why that isn't the case
Best Answer
You have $f(n, n) = \Gamma(n + 1)$ but this by no means implies that $\frac d{dx}f(x, n)\vert_{x = n} = \Gamma'(n + 1)$.
Even if you can extend the arguments of $f$ to a differentiable function $f(x, y)$ such that $f(n, n) = \Gamma(n + 1)$ holds for all $n$ in an open subset of $\Bbb R$, you would get $\Gamma'(n + 1) = \left(\frac \partial{\partial x}f + \frac \partial{\partial y}f\right)\vert_{x, y = n, n}$.