How to find the derivative of the Beta function? From a standard reference book I found that:
${\partial \over \partial x}\mathrm{B}(x, y) = \mathrm{B}(x, y) \left( {\Gamma'(x) \over \Gamma(x)} - {\Gamma'(x + y) \over \Gamma(x + y)} \right)$. 
Currently, I would like to find:
${\partial \over \partial x}\mathrm{B}(1+x, 1-x)$
However, I am unsure how to handle the extra constant inside and am unable to see how the derivative was found. Would this derivative be as simple as just plugging into the equation above? Thanks.
 A: HINT:
$$B(1+x,1-x)=\frac{\Gamma(1+x)\Gamma(1-x)}{\Gamma(2)} \tag 1$$
and 
$$\Gamma(1-x)=\frac{\pi}{\sin (\pi x)\Gamma(x)} \tag 2$$
for $0<x<1$.
SPOILER ALERT  Scroll over the highlighted area to reveal the answer

Using $(2)$ in $(1)$ yields $$\begin{align}B(1+x,1-x)&=\frac{\Gamma(1+x)\pi}{\sin(\pi x)\Gamma(x)}\\\\&=\frac{\pi x}{\sin(\pi x)} \tag 3 \end{align}$$where we used both (i) $\Gamma(2)=1$ and (ii) the Functional Equation $\Gamma(x+1)=x\Gamma(x)$ for the Gamma function.  Now, taking the derivative on both sides of $(3)$ reveals that $$\bbox[5px,border:2px solid #C0A000]{\frac{d B(1+x,1-x)}{dx}=\pi \csc(\pi x)\left(1-(\pi x)\cot (\pi x)\right)}$$And we are done! Note that for $x=0$, we may remove the discontinuity of this expression for the derivative by simply setting the value equal to the limit.  This limit is easily found to be $0$ using standard asymptotic analysis with $\csc(\pi x)\sim \frac{1}{\pi x}$ and $\cot (\pi x)\sim \frac{1-\frac12 (\pi x)^2}{\pi x}$.

A: Let's expand the Beta in terms of Gamma-functions:
$$ B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, $$
so $B(1-x,1+x) = \Gamma(1-x)\Gamma(1+x)/\Gamma(2)$. $\Gamma(2)=1$. Meanwhile,
$$ \Gamma(1-x)\Gamma(1+x) = x \Gamma(x)\Gamma(1-x) = x \frac{\pi}{\sin{\pi x}} $$
using the reflection formula, so
$$ \frac{d}{dx} B(1-x,1+x) = \pi\csc{\pi x} - \pi^2 x \csc{\pi x} \cot{\pi x} = \pi  \left( 1 - \pi x \cot{\pi x} \right) \csc{\pi x}.$$
