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?

Logarithmic differentiation is straightforward:  note that for a differentiable function $f$ whose logarithm is well-defined, $$\frac{d}{dx}\left[\log f(x)\right] = \frac{f'(x)}{f(x)},$$ so we have with the choice $f(x) = \prod_{n=1}^m (x+1/n)$ $$\frac{df}{dx} = \frac{d}{dx}\left[\prod_{n=1}^m \left( x + \frac{1}{n} \right)\right] = \left(\prod_{n=1}^m \left(x + \frac{1}{n}\right)\right)\left( \frac{d}{dx}\left[\sum_{n=1}^m \log \left(x + \frac{1}{n}\right) \right] \right),$$ thus converting the derivative of a product into a sum of derivatives.  Continuing, we find $$\frac{df}{dx} = f(x) \sum_{n=1}^m \frac{1}{x+1/n}.$$  Evaluating at $x = 0$ gives $$f'(0) = \frac{m(m+1)/2}{m!} = \frac{m+1}{2(m-1)!}.$$
