Question Number 129039 by oustmuchiya@gmail.com last updated on 12/Jan/21
$${partially}\:{Differentiate}\:{function} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)=\mathrm{2}\boldsymbol{{x}}^{−\mathrm{2}} \boldsymbol{{y}}+\boldsymbol{{xy}}^{\mathrm{3}} +\frac{\mathrm{2}}{\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}} \\ $$
Answered by liberty last updated on 12/Jan/21
$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{2x}^{−\mathrm{2}} \mathrm{y}+\mathrm{xy}^{\mathrm{3}} +\mathrm{2x}^{−\mathrm{2}} \mathrm{y}^{−\mathrm{1}} \\ $$$$\begin{cases}{\frac{\partial\mathrm{f}}{\partial\mathrm{x}}\:=\:−\mathrm{4x}^{−\mathrm{3}} \mathrm{y}+\mathrm{y}^{\mathrm{3}} −\mathrm{4x}^{−\mathrm{3}} \mathrm{y}^{−\mathrm{1}} }\\{\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\:=\:\mathrm{2x}^{−\mathrm{2}} +\mathrm{3xy}^{\mathrm{2}} −\mathrm{2x}^{−\mathrm{2}} \mathrm{y}^{−\mathrm{2}} \:}\end{cases} \\ $$