Question Number 821 by 123456 last updated on 17/Mar/15
$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({xg}\right)={f}\left({x}\right){g}\left({y}\right) \\ $$$${g}\left({xy}\right)={f}\left({x}\right)+{g}\left({y}\right) \\ $$$$\frac{{d}\left({fg}\right)}{{dx}}=? \\ $$
Answered by prakash jain last updated on 18/Mar/15
$${g}\left(\mathrm{1}\right)={f}\left(\mathrm{1}\right)+{g}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${f}\left({xy}\right)={f}\left({x}\right){g}\left({y}\right)\Rightarrow{f}\left({y}\right)={f}\left(\mathrm{1}\right){g}\left({y}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=\mathrm{0} \\ $$$${g}\left({xy}\right)={g}\left({y}\right) \\ $$$${g}\left({x}\right)={g}\left(\mathrm{1}\right)\:\Rightarrow{g}\left({x}\right)={C}\mathrm{onstant} \\ $$$${f}\left({x}\right){g}\left({x}\right)=\mathrm{0} \\ $$