Derivative of $(x-1)(x-2)(x-3) \cdots (x-10)$ at $x=6$

$P(x)=\prod\limits_{k=1}^{10}(x-k)=(x-6)Q(x)$ where $Q(x)=\prod\limits_{k=1\\k\neq 6}^{10}(x-k)$

$P'(x)=Q(x)+(x-6)Q'(x)$

Thus $P'(6)=Q(6)+0=Q(6)=2880$

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).