Derivative of the binomial $\binom x n$ with respect to $x$ My background is not mathematics and I need to implement (in C++) the derivative of a binomial, with wxMaxima and wolfram.alpha as a helper. So far, the binomial can be written as:
$$\binom x n = \frac 1 {n!}\prod_{k=1}^n (x-k+1)$$
This reduces to a continued convolution. For my specific needs, the binomial needs to be of the form:
$$\binom{\frac{n+1}{2}x+\frac{n-1}{2}}{n}$$
But I also need the derivative of it, which wxMaxima solves as
$$-\frac{1}{2}(n+1) \,\left( \psi_0\left( \frac{(n+1) x-n+1} 2 \right) -\psi_0 \left( \frac{(n+1) \,(x+1) }{2}\right) \right) \,\begin{pmatrix}\frac{( n+1) x+n-1}{2}\\ n
\end{pmatrix}$$
while wolfram goes a bit further and, instead of $\psi_0$ gives $H_n$, which they call harmonic number. (this link). That $\psi$ seems to have quite an involved formula, but $H_n$,as functions.wolfram has it, is a simple $\sum_{k=1}^n 1/k$, which is a lot simpler in terms of C++.
Now, because I have trust issues, I went on to verify the answer given by wolfram, in wxMaxima, for $n=4$. Here's the code:
n:4$
g:diff(binomial((n+1)/2*(x+1)-1,n),x),expand,numer$
h:-(n+1)/2*binomial((n+1)/2*(x+1)-1,n)*(sum(1/k,k,1,(n+1)/2*(x-1))-sum(1/k,k,1,1/2*(n*(x+1)+x-1)));
wxplot2d([g,h],[x,0,1]);

and here's the output of it:
plot
As you can see, they don't match; plotting wxMaxima's derivation is a match, but that involves $\psi$ as an infinite sum. So I'm left wondering what's wrong: is the implementation of the harmonic number? Is the derivation formula? Is it the way I transcribed it?
TL;DR: I need a derivation formula (not the actual code, that's up to me) for the binomial that is (fairly) simple to implement and doesn't take ages to compute, in C++, as the whole function will be called in a bracketed root-finding algorithm. And I'm also using GMP from gmplib dot org (need 10 rep to post more than 2 links).

Following G Cab's excellent post, and modifying the formulas according to my needs, I managed to come up (with a bit of hammering) to this formula:
$$\frac{d}{dx}\binom{\frac{n+1}{2}x+\frac{n-1}{2}}{n}=(-1)^n \frac{n+1}{2} \binom{\frac{n+1}{2}x + \frac{n-1}{2}}{n} \sum_{k=0}^{n-1} \frac{1}{\frac{n+1}{2}x +\frac{n-1}{2}-k}$$
The $(-1)^n$ takes care of odd $n$. Thank you very much everyone that answered.
 A: We can start with the product representation
\begin{align*}
\binom{\frac{n+1}{2}x+\frac{n-1}{2}}{n}&=\frac{1}{n!}\prod_{k=1}^n\left(\frac{n+1}{2}x+\frac{n-1}{2}-k+1\right)\\
&=\frac{1}{2^nn!}\prod_{k=1}^n\left((n+1)x+n+1-2k\right)
\end{align*}
and recall that 
\begin{align*}\frac{d}{dx}\prod_{k=1}^nf_k(x)
=\sum_{j=1}^nf_j^\prime(x)\prod_{{k=1}\atop{k\neq j}}^nf_k(x)
\end{align*}

We obtain
  \begin{align*}
\frac{d}{dx}\binom{\frac{n+1}{2}x+\frac{n-1}{2}}{n}
&=\frac{d}{dx}\left(\frac{1}{2^nn!}\prod_{k=1}^n\left((n+1)x+n+1-2k\right)\right)\\
&=\frac{1}{2^nn!}\sum_{j=1}^n
\left(\frac{d}{dx}\left((n+1)x+n+1-2j\right)\right)
\prod_{{k=1}\atop{k\neq j}}^n\left((n+1)x+n+1-2k\right)\\
&=\frac{n+1}{2^nn!}\sum_{j=1}^n\prod_{{k=1}\atop{k\neq j}}^n
\left((n+1)x+n+1-2k\right)
\end{align*}

A: For the purpose of computing the derivatives, you can also profitably make  use of the expression of the binomials via the Stirling Numbers of 1st kind
$$
\binom y n = \frac{y^{\,\underline {\,n\,} }}{n!} = \frac{1}{n!} \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} (-1)^{n - k} \left[ \begin{gathered}
  n \\ 
  k
\end{gathered}  \right]y^k
$$
Now, to explain about the alternative formulas with $\psi$ and $H$,
consider the binomial expressed in terms of the Gamma function
$$
\binom y n = \frac{\Gamma (y + 1)}{\Gamma (n + 1)\Gamma (y - n + 1)} = \frac 1 {n!} \frac{\Gamma (y + 1)} {\Gamma (y - n + 1)}
$$
then
$$
\begin{gathered}
  \frac d {dy} \binom y n =  - \frac 1 {n!} \frac{\Gamma '(y + 1)\Gamma (y - n + 1) - \Gamma (y + 1)\Gamma '(y - n + 1)}
{\Gamma (y - n + 1)^2} =  \hfill \\
   = \frac 1 {n!} \left( {\frac{\Gamma (y + 1)}{{\Gamma (y - n + 1)}} \frac{{\Gamma '(y - n + 1)}}
{\Gamma (y - n + 1)} - \frac{{\Gamma (y + 1)}}
{\Gamma (y - n + 1)}\frac{\Gamma '(y + 1)}
{\Gamma (y + 1)}} \right) =  \hfill \\
   = \binom y n (\psi _0 (y - n + 1) - \psi _0 (y + 1)) \hfill \\ 
\end{gathered}
$$
Consider instead the binomial expressed in terms of the product,
$$
\binom y n = \frac{y^{\,\underline {\,n\,} }}{n!} = \frac 1 {n!} \prod\limits_{j = 0}^{n - 1} (y - j)
$$
then for the derivative you have
$$
\begin{gathered}
  \frac d {dy} \binom y n = \frac 1 {n!} \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\prod\limits_{\begin{array}{*{20}c}
   {j \ne k}  \\
   {j = 0}  \\
 \end{array} }^{n - 1} {(y - j)} }  = \frac 1 {n!} \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\frac{1}{y - k} \prod\limits_{j = 0}^{n - 1} {\left( {y - j} \right)}  = }  \hfill \\
   = \binom y n \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\frac{1}
{{y - k}}}  = \binom y n \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} \frac 1 {y - n + k} \hfill \\ 
\end{gathered} 
$$
and for computational purposes, this formula is already quite viable,
and I think you do not need to consider the further expansion leading to:
$$
\frac d {dy} \binom y n = \binom y n \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} \frac 1 {y - n + k} = \binom y n \sum_{y - n + 1\, \leqslant \,k\, \leqslant \,y} \frac 1 k = \binom y n (H_y  - H_{y - n})
$$
A: As we know, the series of the binomial coefficients is the binomial series,
$$
F(x,z)=\sum_{n=0}^\infty\binom xn z^n=(1+z)^x
$$
This has the $x$ derivative
$$
\frac{\partial}{\partial x}F(x,z)
=\ln(1+z)\,(1+z)^x
=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}z^k\sum_{m=0}^\infty\binom xm z^m
=\sum_{n=1}^\infty\sum_{k=1}^n\frac{(-1)^{k-1}}{k}\binom x{n-k} z^n
$$
so that
$$
\frac{d}{dx}\binom xn = \sum_{k=1}^n\frac{(-1)^{k-1}}{k}\binom x{n-k}.
$$
