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?
 A: We can write the product as
$$\exp\left(\sum_{n=1}^{2014}\log(x+1/n)\right)$$
A: You are being asked to evaluate the derivative of a polynomial $f(x) = a_0 + a_1 x + \ldots $ when $x = 0$. Since $f'(x) = a_1 + 2a_2 x + \ldots$, the  result will be $a_1$. In your case (putting $N = 2014$) $a_1$ is given by
$$
\sum_{n=1}^N\prod_{\begin{array}{c}m = 1\\m \neq n\end{array}}^N\frac{1}{m} = \sum_{n=1}^N \frac{n}{N!} = \frac{1}{N!}\sum_{n=1}^N n = \frac{1}{N!}\frac{1}{2}N(N+1) = \frac{N+1}{2\cdot(N-1)!}
$$
A: $\frac{d}{dx}\prod_{n=1}^{2014}\left(x+\frac{1}{n}\right) = \sum_\limits{i=1}^{2014} \prod_\limits{n\ne i}(x+\frac 1n)$
evaluating at $x=0$ gives us
$\sum_\limits{i=1}^{2014} \prod_\limits{n\ne i}(\frac 1n)\\\prod_\limits{n\ne i}(\frac 1n) = \frac {i}{2014!}\\
\frac 1{2014!}\sum_\limits{i=1}^{2014}i = \frac {(2015)(2014)}{2(2014!)} =  \frac {2015}{2(2013!)}$
A: 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.
A: 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 $$f'(0) = \frac{m(m+1)/2}{m!} = \frac{m+1}{2(m-1)!}.$$
A: HINT:
Write $\prod_{n=1}^{2014}\left(x+\frac1n\right)=e^{\sum_{n=1}^{2014}\log\left(x+\frac1n\right)}$
Can you proceed now?
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\left.\totald{}{x}\prod_{n = 1}^{2014}\pars{x + {1 \over n}}
\right\vert_{\ x\ =\ 0} = 1 + {1 \over 2} + {1 \over 3} + \cdots +
{1 \over 2014} =
\bbx{\ds{H_{2014}}} \approx \ln\pars{2015} + \gamma \approx 8.1856
\end{align}
A: If expanded: 
$$\frac{d}{dx} \prod_{n=1}^{2014} \left(x+\frac{1}{n}\right) \big\vert_{x=0}=\frac{d}{dx} \prod_{n=1}^{2014} \left(\frac{nx+1}{n}\right) \big\vert_{x=0}= \frac{\frac{d}{dx}\prod_{n=1}^{2014} (nx+1) \big\vert_{x=0}}{\prod_{n=1}^{2014} n}= \frac{\frac{d}{dx} \left[(x+1)(2x+1)(3x+1)\cdots(2014x+1)\right] \big\vert_{x=0}}{2014!}=\frac{\frac{d}{dx} \left[(C_{2014}x^{2014}+C_{2013}x^{2013}+\cdots+C_{2}x^2+ (1+2+\cdots+2014)x+1)\right] \big\vert_{x=0}}{2014!}=\frac{1+2+\cdots+2014}{2014!}=\frac{2015}{2\cdot2013!}$$
