Derivative of a large product I need help computing 
$$
\frac{d}{dx}\prod_{n=1}^{2014}\left(x+\frac{1}{n}\right)\biggr\rvert_{x=0}
$$
The answer provided is $\frac{2015}{2\cdot 2013!}$, however, I do not know how to arrive at this answer. Does anyone have any suggestions?

This big product is hard to work with, but we can turn it into an easier-to-work-with summation by taking advantage of logarithms.  Let $\displaystyle f(x) = \prod_{n=1}^{2014} \left(x + \frac{1}{n} \right)$.  
Chain rule gives $\displaystyle \Big(\ln(f(x)) \Big)' = \frac{f'(x)}{f(x)}$, which means $f'(x) = f(x) \Big( \ln(f(x)) \Big)'$.  
A property of logarithms tells us that $\ln(f(x)) = \displaystyle \sum_{n=1}^{2014} \ln \left( x + \frac{1}{n} \right)$.  The derivative of this is $\displaystyle \sum_{n=1}^{2014} \frac{1}{x + 1/n}$.  Now let's evaluate $f'(x)$ at zero:
$$\displaystyle f'(0) = \left( \prod_{n=1}^{2014}  \frac{1}{n} \right) \left( \sum_{n=1}^{2014} n \right)$$ 
Apply the famous formula for the sum of the first $n$ natural numbers and you arrive at what you're looking for.
