Hello everyone!

I have a question about the derivative of the regularized upper incomplete gamma function. Considering the *gamma function* and the *incomplete gamma function*
\begin{eqnarray}
\Gamma(x)&=&\int_0^\infty{t^{x-1}e^{-t}\mathrm{d}t},\\\
\Gamma(x,\lambda)&=&\int_\lambda^\infty{t^{x-1}e^{-t}\mathrm{d}t},\end{eqnarray}
where the quotient of $\Gamma(x,\lambda)$ divided by $\Gamma(x)$ is usually denoted as *regularized upper incomplete gamma function*
\begin{equation}
Q(x,\lambda)=\frac{\Gamma(x,\lambda)}{\Gamma(x)},
\end{equation}
**there seems to be a precise estimation for the derivative $\frac{\partial Q(x,\lambda)}{\partial x}$, but I cannot prove it.**

With numerical experiment, I found that given $\lambda$ is not small, $\frac{\partial Q(x,\lambda)}{\partial s}$ is almost proportional to
\begin{equation}
\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}=\exp\left(-\left(x\log x-x\log\lambda-x+\lambda\right)\right).
\end{equation}
Denoting
\begin{equation}
I(\lambda)=\int_0^\infty{\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}\mathrm{d}x},
\end{equation}
**the partial derivative $\frac{\partial Q(x,\lambda)}{\partial x}$ seems to have the following approximation**
\begin{equation}
\frac{\partial Q(x,\lambda)}{\partial x}
\approx\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}/I(\lambda).
\end{equation}

I give the comparison of these two formulas from the numerical experiments with Matlab in following figures, where the *blue solid line* and *green dotted line* represent $\frac{\partial Q(x,\lambda)}{\partial x}$ and $\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}/I(\lambda)$ respectively.

From these figures, it's obvious that as the increase of $\lambda$, the two lines approach to be equal. However, I cannot prove such a relationship due to the complexity of the derivative $\frac{\partial Q(x,\lambda)}{\partial x}$.

Could you help me with some comments about this question? Any suggestions will be helpful and thank you very much!