Question Number 26947 by prakash jain last updated on 31/Dec/17
$$\frac{{d}}{{dx}}\mathrm{ln}\:\left(\Gamma\left({x}+\mathrm{1}\right)\right)=? \\ $$
Commented by abdo imad last updated on 31/Dec/17
$${we}\:{know}\:{that}\:\Gamma\left({x}+\mathrm{1}\right)={x}\:\Gamma\left({x}\right)\:{with}\:{x}>\mathrm{0} \\ $$$$\Rightarrow{ln}\left(\Gamma\left({x}+\mathrm{1}\right)\right)\:=\:\:{lnx}\:+{ln}\left(\Gamma\left({x}\right)\right) \\ $$$$\Rightarrow\frac{{d}}{{dx}}{ln}\left(\Gamma\left({x}+\mathrm{1}\right)\right)=\:\frac{\mathrm{1}}{{x}}+\:\frac{\Gamma^{,} \left({x}\right)}{\Gamma\left({x}\right)}\:. \\ $$
Answered by Femmy last updated on 31/Dec/17
$$\mathrm{ans} \\ $$$$\mathrm{ln}\Gamma+\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right) \\ $$$$\mathrm{1}/\Gamma+\mathrm{1}/\left(\mathrm{x}+\mathrm{1}\right) \\ $$$$ \\ $$
Commented by prakash jain last updated on 31/Dec/17
$$\mathrm{Actually}\:\mathrm{I}\:\mathrm{meant}\:\Gamma\:\mathrm{as}\:\mathrm{gamma}\:\mathrm{function} \\ $$
Commented by Femmy last updated on 31/Dec/17
$$\mathrm{okay}\:\mathrm{sir} \\ $$$$ \\ $$