# Derivative of Incomplete Gamma Function

For the following upper incomplete Gamma function:

$$\Gamma(1+d,A-c \ln x)=\int_{A-c\ln x}^{\infty}t^{(1+d)-1}e^{-t}dt$$ I am trying to calculate the derivative of $Γ$ with respect to $x$. In general it holds that: $$\frac{\mathrm{d} }{\mathrm{d} x} \Gamma(s,x)=-x^{s-1}e^{-x}$$ After my calculations I ended up with: $$\frac{\mathrm{d} }{\mathrm{d} x} \Gamma(1+d,A-c \ln x)=c e^{-A}x^{c-1}(A-c\ln x)^d$$ but the author says that the correct answer is rather: $$\frac{\mathrm{d} }{\mathrm{d} x} \Gamma(1+d,A-c \ln x)=-e^{-A}x^{c-1}(A-c\ln x)^d$$ But how is this correct? By applying the chain rule it should be: $$\frac{\mathrm{d} }{\mathrm{d} x} \Gamma(1+d,A-c \ln x)=\left[ -(A-c\ln x)^de^{A-c \ln x} \right]\frac{\partial{}}{\partial{x}}(A-c \ln x)$$ right?

Thanks!

Hint. We have $$\frac{\partial}{\partial x} Γ(s,x)=-x^{s-1}e^{-x}$$ then, by the chain rule, we get \begin{align} \frac{d}{d x}\left(Γ(1+d,A-c \ln x)\right)&=\frac{-c}{x}\cdot\left.\frac{\partial}{\partial t} Γ(s,t)\right|_{(s,t) =(1+d,A-c \ln x)} \\\\&=\frac{c}{x}\cdot(A-c \ln x)^de^{-(A-c \ln x)} \\\\&=c\:x^{c-1}e^{-A}(A-c \ln x)^d. \end{align}
• Thank you for your answer. Are you sure that there it is $x^c$ and not $x^{c-1}$? It would be of great assistance if you could be a bit more detailed on why this is the result. I would like to understand my fault on applying the chain rule :) – Mitscaype Jun 25 '16 at 17:25
• I have a question. Is $Γ_x(1+d,A-c \ln x)$ meaning $\frac{d}{dx}\left(Γ(1+d,A-c \ln x)\right)$ or is it meaning $\left.\left(\frac{\partial}{\partial t}Γ(1+d,t)\right)\right|_{t=A-c \ln x}$? – Olivier Oloa Jun 25 '16 at 17:30
• It is $\frac{d}{dx} \left( \Gamma(1+d,A-c\ln x) \right)$. – Mitscaype Jun 25 '16 at 17:37