Question Number 60426 by cesar.marval.larez@gmail.com last updated on 20/May/19
$${the}\:{function}\:{is}\:{considered}\: \\ $$$${f}\left({x},{y}\right)={e}^{{xy}} +\frac{{x}}{{y}}+{sen}\left(\left(\mathrm{2}{x}+\mathrm{3}{y}\right)\pi\right)\:{Calcule}: \\ $$$$\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{y}},\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} },\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}.\:\:\:{f}_{{x}} \left(\mathrm{0},\mathrm{1}\right),{f}_{{y}} \left(\mathrm{2},−\mathrm{1}\right),\:{f}_{{xx}} \left(\mathrm{0},\mathrm{1}\right),{f}_{{xy}} \left(\mathrm{2},−\mathrm{1}\right) \\ $$
Commented by kaivan.ahmadi last updated on 21/May/19
$$\frac{\partial{f}}{\partial{x}}={ye}^{{xy}} +\frac{\mathrm{1}}{{y}}+\mathrm{2}\pi{cos}\left(\left(\mathrm{2}{x}+\mathrm{3}{y}\right)\pi\right) \\ $$$$\left.\frac{\partial{f}}{\partial{y}}={xe}^{{xy}} −\frac{{x}}{{y}^{\mathrm{2}} }+\mathrm{3}\pi{cos}\left(\mathrm{2}{x}+\mathrm{3}{y}\right)\pi\right) \\ $$$$\left.\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }=\frac{\partial}{\partial{x}}\left(\frac{\partial{f}}{\partial{x}}\right)={y}^{\mathrm{2}} {e}^{{xy}} −\mathrm{4}\pi^{\mathrm{2}} {sin}\left(\mathrm{2}{x}+\mathrm{3}{y}\right)\pi\right) \\ $$$$\left.\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}=\frac{\partial}{\partial{x}}\left(\frac{\partial{f}}{\partial{y}}\right)={xye}^{{xy}} −\frac{\mathrm{1}}{{y}^{\mathrm{2}} }−\mathrm{6}\pi^{\mathrm{2}} {sin}\left(\mathrm{2}{x}+\mathrm{3}{y}\right)\pi\right) \\ $$$${f}_{{x}} =\frac{\partial{f}}{\partial{x}}\:\:\:,… \\ $$$${f}_{{x}} \left(\mathrm{0},\mathrm{1}\right)={e}^{\mathrm{0}} +\mathrm{1}+\mathrm{2}\pi{cos}\mathrm{3}\pi=\mathrm{2}−\mathrm{2}\pi \\ $$$${f}_{{y}} \left(\mathrm{2},−\mathrm{1}\right)=\mathrm{2}{e}^{−\mathrm{2}} +\mathrm{3}\pi{cos}\pi=\frac{\mathrm{2}}{{e}^{\mathrm{2}} }−\mathrm{3}\pi \\ $$$${f}_{{xx}} \left(\mathrm{0},\mathrm{1}\right)={e}^{\mathrm{0}} −\mathrm{4}\pi^{\mathrm{2}} {sin}\mathrm{3}\pi=\mathrm{1} \\ $$$${f}_{{xy}} \left(\mathrm{2},−\mathrm{1}\right)=−\mathrm{2}{e}^{−\mathrm{2}} +\mathrm{1}−\mathrm{6}\pi^{\mathrm{2}} {sin}\pi=\frac{−\mathrm{2}}{{e}^{\mathrm{2}} }+\mathrm{1} \\ $$
Commented by cesar.marval.larez@gmail.com last updated on 21/May/19
$${thanks}\:{sir}\:{for}\:{the}\:{solution} \\ $$