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.
Best Answer
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 $0<x<1$.
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