Derivative of the binomial $\binom x n$ with respect to $x$

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*}

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: $$ \frac d {dy} \binom y n = \binom y n \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} \frac 1 {y - n + k} = \binom y n \sum_{y - n + 1\, \leqslant \,k\, \leqslant \,y} \frac 1 k = \binom y n (H_y - H_{y - n}) $$

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 $$ \frac{d}{dx}\binom xn = \sum_{k=1}^n\frac{(-1)^{k-1}}{k}\binom x{n-k}. $$