Derivative of Gamma function In Computing the integral of $\log(\sin x)$, user17762 provided a solution which requires differentiating $\displaystyle \frac{\Gamma(2z+1)}{4^z\Gamma^2(z+1)}\frac{\pi}{2}$ with respect to $z$. How is this done?
I had actually got $\displaystyle\int_0^{\pi/2}\sin^{2z}(x)dx = \frac{\Gamma(z+\frac{1}{2})}{\Gamma(z+1)}\frac{\sqrt\pi}{2}$ instead. But I am guessing they are equivalent and differentiating them would use the same technique. 

Regarding the two expressions and your doubt about their equality: 
The equality of $\displaystyle \frac{\Gamma(2z+1)}{4^z\Gamma^2(z+1)}\frac{\pi}{2}$ and $\displaystyle \frac{\Gamma(z+\frac{1}{2})}{\Gamma(z+1)}\frac{\sqrt\pi}{2}$ can be shown by using the fact that $\Gamma(z)\Gamma(z+\frac12)=2^{1-2z}\sqrt{\pi}\Gamma(2z)$ (see wiki):
$$\Gamma(2z+1)=\Gamma\left(2\left(z+\frac12\right)\right) = \frac{\Gamma\left(z+\frac12\right)\Gamma\left(z+1\right)}{2^{-2z}\sqrt{\pi}}$$
Thus,
$$\frac{\Gamma(2z+1)}{4^z\Gamma^2(z+1)}\frac{\pi}{2}=\frac{\Gamma\left(z+\frac12\right)\Gamma\left(z+1\right)}{2^{-2z}\sqrt{\pi}4^z\Gamma^2(z+1)}\frac{\pi}2=\frac{\Gamma(z+\frac{1}{2})}{\Gamma(z+1)}\frac{\sqrt\pi}{2}$$
