Derivative $ \frac{ d^2}{d^2 x} \frac{\Gamma(x+1) }{\Gamma(x+3)}$ I know this may seem like a really low level/silly question, I apologize in advance.  
I do not know how to differentiate the gamma,beta,digamma function, for ex:
$$
\frac{ d^2}{d^2 x} \frac{\Gamma(x+1) }{\Gamma(x+3)}
$$
how would we do this?  Thank you for the help.
I am not sure what to do.  I know 
$$
\frac{d}{d x}\frac{ \Gamma'(x+1)\Gamma(x+3)-\Gamma(x+1)\Gamma'(x+3)}{\big(\Gamma(x+3)^2\big)}=...
$$
I can do the second derivative too, I just am not familiar with how to do this and I notice it a lot.  THanks
 A: You don't need anything that complicated.  Just show that by integration by parts
$$\frac{\Gamma(x+1)}{\Gamma(x+3)} = \frac{1}{(x+1)(x+2)},$$
and then differentiate.
Yes, this works for any real $x>0$.
Detail: From the definition of the Gamma function,
$$\forall x \geq 0: \Gamma(x+1)=\int_0^\infty t^x e^{-t} dt = [-t^x e^{-t}]_0^\infty + x \int_0^\infty t^{x-1} e^{-t} dt = x \Gamma(x).$$
A: By, one of the many, definition the digamma function is:
\begin{align}
\psi(x) = \partial_{x} \ \ln\Gamma(x) = \frac{\Gamma^{'}(x)}{\Gamma(x)}.
\end{align}
Now consider
\begin{align}
f(x,y,t) &= \ \frac{\Gamma(ax+by)}{\Gamma(cx+dt)}
\end{align}
for which the following derivatives are seen:
\begin{align}
\partial_{x} \ f(x,y,t) &= \frac{1}{\Gamma^{2}(cx+dt)} \left[ a \Gamma^{'}(ax+by) \Gamma(cx+dt) - c \Gamma(ax+by) \Gamma^{'}(cx+dt) \right] \\
&= \frac{\Gamma(ax+by)}{\Gamma(cx+dt)} \left[ a \psi(ax+by) - c \psi(cx+dt) \right] \\
&= f(x,y,t) \left[ a \psi(ax+by) - c \psi(cx+dt) \right].
\end{align}
Differentiation with respect to $y$ is
\begin{align}
\partial_{y} \ f(x,y,t) = b \ f(x,y,t) \ \psi(ax+by)
\end{align}
and differentiation with respect to $t$ is
\begin{align}
\partial_{t} \ f(x,y,t) &= - d \ f(x,y,t) \ \psi(cx+dt).
\end{align}
In the case proposed the following is seen.
\begin{align}
\partial_{x}^{2} \ \frac{\Gamma(x+a)}{\Gamma(x+a+2)} &= \partial_{x}^{2} \ \frac{\Gamma(x+a)}{(x+a+1)(x+a)\Gamma(x+a)} \\
&= \partial_{x}^{2} \ \frac{1}{(x+a+1)(x+a)} 
= \partial_{x}^{2} \left[ \frac{1}{x+a} - \frac{1}{x+a+1} \right] \\
&= \frac{1}{(x+a)^{2}} - \frac{1}{(x+a+1)^{2}} = \frac{2x + 2a+1}{(x+a)^{2}(x+a+1)^{2}} \\
&= (2x+2a+1) \left( \frac{\Gamma(x+a)}{\Gamma(x+a+2)} \right)^{2}.
\end{align}
