\[\frac{\partial J_{\pm\nu}\left(z\right)}{\partial\nu}=\pm J_{\pm\nu}\left(z%
\right)\ln\left(\tfrac{1}{2}z\right)\mp(\tfrac{1}{2}z)^{\pm\nu}\sum_{k=0}^{%
\infty}(-1)^{k}\frac{\psi\left(k+1\pm\nu\right)}{\Gamma\left(k+1\pm\nu\right)}%
\frac{(\tfrac{1}{4}z^{2})^{k}}{k!},\]