\[\frac{\mathrm{d}}{\mathrm{d}z}\left(z^{-\nu}\mathbf{L}_{\nu}\left(z\right)%
\right)=\frac{2^{-\nu}}{\sqrt{\pi}\Gamma\left(\nu+\tfrac{3}{2}\right)}+z^{-\nu%
}\mathbf{L}_{\nu+1}\left(z\right).\]