Given $f(x)=x(x-1)(x-2)...(x-10)$ what is the derivative $f'(0)$? 
$$f: \Bbb R \to \Bbb R; x \mapsto f(x)=x(x-1)(x-2)\cdots(x-10)$$
  Evaluate $f'(0)$!

I've tried to set the factors apart, but I only know that $(fg)'=f'g+fg'$.
I don't know how I should apply that rule for any $n$ amount of factors. I also thought of actually doing the multiplication, but I don't know what shortcut I should use, and multiplicating one after the other takes extremely long.
 A: Hint: A polynomial is always an entire function, and in a neighbourhood of the origin:
$$ x(x-1)\cdot\ldots\cdot(x-10) = 10!\,x+O(x^2) $$
hence
$$ \frac{d}{dx}\left. x(x-1)\cdot\ldots\cdot(x-10)\right|_{x=0} = \color{red}{10!}. $$
A: Welcome to Math SE. To format your question you can use LaTeX. 
Now coming to your question, the derivative of a product is just the sum of the $n$ products where only one of the members is differentiated. In your case
$((x)... (x-10))' = [(x-1)...(x-10)]+[x(x-2)...(x-10)]+...+[x(x-1)...(x-9)]$
Note that all but the first expression will evaluate to $0$ at $x=0$ since you're multiplying by 0 ($x$) so $f'(0)=(-1)(-2)...(-10)=10!$
A: Let $g(x) = x$ and $h(x) = (x-1)\cdots (x-10)$. Then $f'(x) = g(x) h'(x) + g'(x)h(x)$. Since $g(0)=0$ and $g'(0)=1$, we have
$$f'(0) = h(0) = (-1)(-2)\cdots (-10) = 10!$$
A: Apply the definition:
$$
f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=
\lim_{x\to0}\,(x-1)(x-2)\dots(x-10)=10!
$$
More generally, if $f(x)=x(x-1)\dots(x-n)$, the same argument shows
$$
f'(0)=(-1)^n\cdot n!
$$
A: There is an $n$-term version of the product rule that is definitely worth knowing about.  You'll see the pattern from the $n = 3$ case.  If $f(x) = f_1(x) f_2(x) f_3(x)$, then
\begin{equation}
f'(x) = f_1'(x) f_2(x) f_3(x) + f_1(x) f_2'(x) f_3(x) + f_1(x) f_2(x) f_3'(x).
\end{equation}
(Do you see what the formula would be for a product of four functions?)
In your case, $f'(x)$ is a sum of $11$ terms, and all but one of those terms vanish when you plug in $x = 0$.
Richard Feynman made a big deal about the usefulness of this $n$-term product rule in The Feynman Tips on Physics.
A: More generally, if
$f(x)
=\prod_{k=1}^n (x-a_k)^{b_k}
$,
then
$\ln f(x)
=\sum_{k=1}^n b_k \ln(x-a_k)
$.
Differentiating,
$\dfrac{f'(x)}{f(x)}
=\sum_{k=1}^n \dfrac{b_k}{x-a_k}
$,
so
$f'(x)
=f(x)\sum_{k=1}^n \dfrac{b_k}{x-a_k}
$.
Setting $x=0$,
$\begin{array}\\
f'(0)
&=f(0)\sum_{k=1}^n \dfrac{b_k}{-a_k}\\
&=\prod_{j=1}^n (-a_j)^{b_j}\sum_{k=1}^n \dfrac{b_k}{-a_k}\\
&=\sum_{k=1}^n b_k(-a_k)^{b_k-1}\prod_{j=1, j \ne k}^n (-a_j)^{b_j}\\
\end{array}
$
If $a_k = k-1$
and $b_k = 1$
as in your case,
$\begin{array}\\
f'(0)
&=\sum_{k=1}^n \prod_{j=1, j \ne k}^n (-j+1)\\
&=(-1)^{n-1}\sum_{k=1}^n \prod_{j=1, j \ne k}^n (j-1)\\
&=(-1)^{n-1}\left(\prod_{j=2}^n (j-1)+\sum_{k=2}^n \prod_{j=1, j \ne k}^n (j-1)\right)\\
&=(-1)^{n-1}(n-1)!
\qquad\text{since all terms with }k\ge 2 \text{ have j=1 so are zero}\\
\end{array}
$
