Question Number 196793 by MrGHK last updated on 31/Aug/23
Answered by Mathspace last updated on 02/Sep/23
$${B}\left({x},{y}\right)=\frac{\Gamma\left({x}\right)\Gamma\left({y}\right)}{\Gamma\left({x}+{y}\right)} \\ $$$$\Rightarrow\frac{\partial{B}}{\partial{x}}\left({x},{y}\right)=\Gamma\left({y}\right).\frac{\Gamma^{'} \left({x}\right)\Gamma\left({x}+{y}\right)−\Gamma\left({x}\right)\Gamma^{'} \left({x}+{y}\right)}{\Gamma^{\mathrm{2}} \left({x}+{y}\right)} \\ $$$$=\Gamma\left({y}\right)\frac{\Psi\left({x}\right)\Gamma\left({x}\right)\Gamma\left({x}+{y}\right)−\Gamma\left({x}\right)\Psi\left({x}+{y}\right)\Gamma\left({x}+{y}\right)}{\Gamma^{\mathrm{2}} \left({x}+{y}\right)} \\ $$$$=\Gamma\left({y}\right)\frac{\Psi\left({x}\right)\Gamma\left({x}\right)−\Gamma\left({x}\right)\Psi\left({x}+{y}\right)}{\Gamma\left({x}+{y}\right)} \\ $$$$=\frac{\Gamma\left({y}\right)\Gamma\left({x}\right)}{\Gamma\left({x}+{y}\right)}\left\{\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\right\}\Rightarrow \\ $$$$\frac{\partial{B}}{\partial{x}}\left({x},{y}\right) \\ $$$$={B}\left({x},{y}\right)\left\{\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\right\}\:\Rightarrow \\ $$$${and} \\ $$$$\frac{\partial^{\mathrm{2}} {B}}{\partial{y}\partial{x}}\left({x},{y}\right) \\ $$$$=\frac{\partial{B}}{\partial{y}}\left(\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\right) \\ $$$$+{B}\left({x},{y}\right)\left\{−\Psi^{'} \left({x}+{y}\right)\right\} \\ $$$$={B}\left({x},{y}\right)\left(\Psi\left({y}\right)−\Psi\left({x}+{y}\right)\right)\left(\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\right. \\ $$$$−{B}\left({x},{y}\right)\Psi^{'} \left({x}+{y}\right) \\ $$$$={B}\left({x},{y}\right)\left\{\left(\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\left(\Psi\left({y}\right)−\Psi\left({x}+{y}\right)−\Psi^{'} \left({x}+{y}\right)\right\}\right.\right. \\ $$
Commented by MrGHK last updated on 02/Sep/23
$$\boldsymbol{{G}}{reat}\:{solution} \\ $$