I want to find the following derivative:
$\frac{\partial\;{\Gamma({1+\frac{\alpha}{x}})}}{\partial{x}}$, where $\Gamma(.)$ is the gamma function and $\alpha$ is a constant. If it matters, $0\leq\alpha\leq 2$ and typically $x>2$. I know that:
$\frac{\partial\;\Gamma(x)}{\partial x}=\psi(x).\Gamma(x)$,
where $\psi(x)$ is the digamma function. However, here we have $f(x)=1+\frac{\alpha}{x}$ as an argument rather than $x$. My first inclination was to apply the chain rule as follows:
$\frac{\partial\;{\Gamma({1+\frac{\alpha}{x}})}}{\partial{x}}=\frac{\partial\;{\Gamma({1+\frac{\alpha}{x}})}}{\partial{({1+\frac{\alpha}{x}})}}.\frac{\partial\;({1+\frac{\alpha}{x}})}{\partial x}=\psi({1+\frac{\alpha}{x}}).\Gamma({1+\frac{\alpha}{x}}).(-\frac{\alpha}{x^2})$,
But I am not sure we can apply the chain rule in the context of the gamma function.
Is the above correct? if it is not, how do we find the derivative of $\Gamma(f(x))$ with respect to $x$? Thanks.