Determine the number of values of $c$ for which the equation has a root of multiplicity $2$. 
Question:
Let $c$ be a fixed real number. Show that a root of the equation
$x(x+1)(x+2)\cdots(x+2009)=c$
can have multiplicity at most $2$. Determine the number of values of $c$ for which the equation has a root of multiplicity $2$.


Solution that I got from the handout:
If $c=0$, we have $2010$ distinct roots of multiplicity $1$. So we can consider $c\ne 0$ and roots $\notin\{-2009,-2008,...,0\}$
Let $f(x)=\prod_{k=0}^{2009}(x+k)-c$
$f'(x)=\prod_{k=0}^{2009}(x+k)\sum_{k=0}^{2009}\frac 1{x+k}$ $\forall x\notin\{-2009,-2008,...,0\}$
$f"(x)=f'(x)\sum_{k=0}^{2009}\frac 1{x+k}$ $-\prod_{k=0}^{2009}(x+k)\sum_{k=0}^{2009}\frac 1{(x+k)^2}$ $\forall x\notin\{-2009,-2008,...,0\}$
If $r$ is a root of multiplicity $>2$, then $f(r)=f'(r)=f"(r)=0$. Then $\prod_{k=0}^{2009}(r+k)\sum_{k=0}^{2009}\frac 1{(r+k)^2}=0$, and so $c\sum_{k=0}^{2009}\frac 1{(r+k)^2}=0$ impossible. The number of double roots of $f(x)$ is the number of roots of $f'(x)$ and so we have $2009$ such roots.
It's then easy to see that each root $r$ has a matching root $r'=-2009-r$ for which we have the same value of $c$ and that no three roots give the same value of $c$ (previous result).
Hence the result ${1005}$ values of $c$
My question is that how did they get $f'(x)=\prod_{k=0}^{2009}(x+k)\sum_{k=0}^{2009}\frac 1{x+k}$


My question is that how did they get $f'(x)=\prod_{k=0}^{2009}(x+k)\sum_{k=0}^{2009}\frac 1{x+k}$

Simplest possible example to demonstrate:
$$
\begin{align}
{d \over dx} (x+1)(x+2) 
&= (x+1) + (x+2)\\
&= (x+1)(x+2)/(x+2) + (x+1)(x+2)/(x+1)\\
&= (x+1)(x+2)\left({1\over x+2}+{1\over x+1}\right)\\
&= \prod(x+k)\sum 1/(x+k)\\
\end{align}
$$
