Question Number 146085 by mathmax by abdo last updated on 10/Jul/21
$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}−\sqrt{\mathrm{x}+\mathrm{2y}} \\ $$$$\left.\mathrm{1}\right)\mathrm{condition}\:\mathrm{on}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{to}\:\mathrm{have}\:\mathrm{f}\:\mathrm{symetric} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\frac{\partial\mathrm{f}}{\partial\mathrm{x}}\:,\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\:,\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{x}\partial\mathrm{y}}\:,\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{y}\partial\mathrm{x}} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{find}\:\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial^{\mathrm{2}} \mathrm{x}}\:\mathrm{and}\:\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial^{\mathrm{2}} \mathrm{y}} \\ $$
Commented by mathmax by abdo last updated on 10/Jul/21
$$\left.\mathrm{3}\right)\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{x}^{\mathrm{2}} }\:\mathrm{and}\:\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{y}^{\mathrm{2}} } \\ $$