Question Number 7939 by tawakalitu last updated on 24/Sep/16
$${f}\left({x},\:{y}\right)\:=\:{xy}^{\mathrm{3}} \:+\:\mathrm{5}{xy}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1} \\ $$$${find}:\:\:{f}_{{x}} \:,\:{f}_{{y}} \:,\:{f}_{{xx}} \:,\:{f}_{{yy}} \:,\:{f}_{{xy}} \:,\:{f}_{{yx}} \\ $$
Commented by Rasheed Soomro last updated on 25/Sep/16
$${What}'{s}\:{meant}\:{by}\:{f}_{{xy}} \:,{f}_{{x}} \:{etc}? \\ $$
Commented by tawakalitu last updated on 25/Sep/16
$${Is}\:{partial}\:{derivative} \\ $$
Commented by tawakalitu last updated on 25/Sep/16
$${or}\:\:{f}\left({x}\right),\:{f}\left({xy}\right),\:{f}\left({xx}\right)\:…… \\ $$
Answered by prakash jain last updated on 02/Oct/16
$${f}\left({x},\:{y}\right)\:=\:{xy}^{\mathrm{3}} \:+\:\mathrm{5}{xy}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1} \\ $$$${f}_{{x}} ={y}^{\mathrm{3}} +\mathrm{5}{y}^{\mathrm{2}} +\mathrm{2} \\ $$$${f}_{{y}} =\mathrm{3}{xy}^{\mathrm{2}} +\mathrm{10}{xy} \\ $$$${f}_{{xx}} =\mathrm{0} \\ $$$${f}_{{yy}} =\mathrm{6}{xy}+\mathrm{10}{x} \\ $$$$\frac{\partial}{\partial{y}}\left(\frac{\partial{f}}{\partial{x}}\right)=\mathrm{3}{y}^{\mathrm{2}} +\mathrm{10}{x} \\ $$$$\frac{\partial}{\partial{x}}\left(\frac{\partial{f}}{\partial{y}}\right)=\mathrm{3}{y}^{\mathrm{2}} +\mathrm{10}{y} \\ $$