Use the product rule and:
$\displaystyle \frac{d}{dx}\left(\sum_{k=0}^{2011} \frac{x^k}{k!}\right)=\sum_{k=0}^{2011} \frac{d}{dx}\left(\frac{x^k}{k!}\right)$
$\displaystyle \frac{kx^{k-1}}{k!}$ is correct, however, you can use your definition of a factorial to cancel a factor of k giving $\displaystyle \dfrac{x^{k-1}}{(k-1)!}$
But don't forget your other term and use the product rule