Question Number 132477 by KZ last updated on 14/Feb/21
$${define} \\ $$$${f}\left({x}.{y}\right)= \\ $$$$\left.\left\{\frac{\mathrm{xy}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)\:}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:{if}\:\left({x}.{y}\right)\neq\right)\mathrm{0}.\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:\left({x}.{y}\right)=\left(\mathrm{0}.\mathrm{0}\right) \\ $$$$ \\ $$$${show}\:{that}\:{f},\frac{\partial{f}}{\partial{x}}\:{and}\:\frac{\partial{f}}{\partial{y}\:\:}\:{are}\: \\ $$$${continuous}\:{on}\:\mathbb{R}^{\mathrm{2}} \\ $$$${show}\:{that}\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}\:{and}\:\frac{\partial^{\mathrm{2}} {f}}{\partial{y}\partial{x}\:}\:{exist}\: \\ $$$${on}\mathbb{R}^{\mathrm{2}} \:{and}\:{are}\:{continuous}\:{on}\: \\ $$$$\mathbb{R}^{\mathrm{2}} \backslash\left\{\mathrm{0}.\mathrm{0}\right\} \\ $$$${show}\:{that}\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}=\mathrm{1}\neq−\mathrm{1}=\frac{\partial^{\mathrm{2}} {f}}{\partial{y}\partial{x}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by guyyy last updated on 20/Feb/21
