I wish to compute the derivative of the upper incomplete gamma function \begin{equation} \Gamma(s,x) = \int_{x}^\infty t^{s-1}e^{-t} \, dt . \end{equation} Wikipedia states of the derivative of $\Gamma(s,x)$ as \begin{equation} \frac{\partial \Gamma(s,x)}{\partial s} = \ln x \Gamma(s,x) + x \Gamma(3,s,x). \end{equation} where \begin{equation} \Gamma(K,s,x) = G_{K-1,K}^{K,0} \left(\frac{0,0,\cdots,0}{s-1,-1,\cdots,-1}\mid x\right) \end{equation} is the Meijer G function. My question is, if I evaluate this equation at x=0, I should get the psi function but instead I get $-\infty$. Could someone please confirm how to use this, and also clarify if the first term is $\ln (x) \Gamma(s,x)$ or $\ln (x \Gamma(s,x))$.
Thanks.