TITLE: Derivative of $(x-1)(x-2)(x-3) \cdots (x-10)$ at $x=6$
QUESTION [11 upvotes]: I can see that by applying the product rule, all resulting expressions containing $(x-6)$ would become zero, hence leaving $(1)(x-1)(x-2)(x-3)(x-4)(x-5)(x-7)(x-8)(x-9)(x-10)$, which, at $x=6$, is $2880$. However, I'm not sure how to formally write out a concise method for finding this solution.

REPLY [2 votes]: Well, it depends on which properties you will take as given. Let's just say that from the usual product rule ($(fg)'=f'g+fg'$) you can obtain
$$ (fgh)'=f'(gh)+f(gh)'=f'gh+fg'h+fgh',$$ and so on. In a general case, you could write
$$\left(\prod_{k=1}^n f_k \right)'=\sum_{k=1}^n f_k' \cdot \prod_{j\not=k} f_j.$$
You may try from there (in your case, $f_k=x-k$ and n=10).