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Question Number 111974 by mohammad17 last updated on 05/Sep/20

Answered by mathmax by abdo last updated on 05/Sep/20

$$1)\mathrm{f}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2\arctan \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\Rightarrow \frac{\partial \mathrm{f}}{\partial \mathrm{x}}(\mathrm{x},\mathrm{y})=2\mathrm{x}\arctan \left(\frac{\mathrm{y}}{\mathrm{x}}\right)+{\mathrm{x}}^2\times \left(\frac{-\mathrm{y}}{{\mathrm{x}}^2(1+\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2})}\right)$$$$=2\mathrm{x}\arctan \left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\frac{{\mathrm{yx}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}\Rightarrow \frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{x}}^2}(\mathrm{x},\mathrm{y})=2\arctan \left(\frac{\mathrm{y}}{\mathrm{x}}\right)$$$$+2\mathrm{x}.\frac{-\mathrm{y}}{{\mathrm{x}}^2(1+\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2})}-\mathrm{y}.\frac{2\mathrm{x}({\mathrm{x}}^2+{\mathrm{y}}^2)-2\mathrm{x}.{\mathrm{x}}^2}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}$$$$=2\arctan \left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\frac{2\mathrm{xy}}{{\mathrm{x}}^2+{\mathrm{y}}^2}-\mathrm{y}.\frac{{2\mathrm{xy}}^2}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}\Rightarrow$$$${\mathrm{f}}_{\mathrm{xx}}=2\arctan \left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\frac{2\mathrm{xy}}{{\mathrm{x}}^2+{\mathrm{y}}^2}-\frac{{2\mathrm{xy}}^3}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}$$

Answered by mathmax by abdo last updated on 05/Sep/20

$$\mathrm{we}\mathrm{have}\mathrm{f}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2\arctan \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\Rightarrow \frac{\partial \mathrm{f}}{\partial \mathrm{y}}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2\frac{1}{\mathrm{x}(1+\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2})}$$$$=\frac{\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{y}}^2}.{\mathrm{x}}^2=\frac{{\mathrm{x}}^3}{{\mathrm{x}}^2+{\mathrm{y}}^2}\Rightarrow \frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{y}}^2}(\mathrm{x},\mathrm{y})=-{\mathrm{x}}^3.\frac{2\mathrm{y}}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}\Rightarrow$$$${\mathrm{f}}_{\mathrm{yy}}=-\frac{{2\mathrm{x}}^3\mathrm{y}}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}$$

Commented by mathmax by abdo last updated on 05/Sep/20

$$\mathrm{we}\mathrm{have}\frac{\partial \mathrm{f}}{\partial \mathrm{y}}(\mathrm{x},\mathrm{y})=\frac{{\mathrm{x}}^3}{{\mathrm{x}}^2+{\mathrm{y}}^2}\Rightarrow \frac{\partial }{\partial \mathrm{x}}\left(\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\right)(...)=\frac{{3\mathrm{x}}^2({\mathrm{x}}^2+{\mathrm{y}}^2)-2\mathrm{x}.{\mathrm{x}}^3}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}$$$$=\frac{{\mathrm{x}}^4+{3\mathrm{x}}^2{\mathrm{y}}^2}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}\Rightarrow {\mathrm{f}}_{\mathrm{xy}}=\frac{{\mathrm{x}}^4+{3\mathrm{x}}^2{\mathrm{y}}^2}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}$$

Answered by mathmax by abdo last updated on 06/Sep/20

$$2)\mathrm{f}(\mathrm{x},\mathrm{y})=\ln \left(\mathrm{xy}\right)+\tan \left(\mathrm{xy}\right)\Rightarrow \frac{\partial \mathrm{f}}{\partial \mathrm{x}}(\mathrm{x},\mathrm{y})=\frac{1}{\mathrm{x}}+\mathrm{y}(1+{\tan }^2\left(\mathrm{xy}\right))$$$$\Rightarrow \frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{x}}^2}(\mathrm{x},\mathrm{y})=-\frac{1}{{\mathrm{x}}^2}+\mathrm{y}(2\mathrm{ytan}\left(\mathrm{xy}\right)(1+{\tan }^2\left(\mathrm{xy}\right))$$$$=-\frac{1}{{\mathrm{x}}^2}+{2\mathrm{y}}^2\tan \left(\mathrm{xy}\right)(1+{\tan }^2\left(\mathrm{xy}\right))={\mathrm{f}}_{\mathrm{xx}}$$$$\mathrm{we}\mathrm{have}\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{f}(\mathrm{y},\mathrm{x})\Rightarrow {\mathrm{f}}_{\mathrm{yy}}=-\frac{1}{{\mathrm{y}}^2}+{2\mathrm{x}}^2\tan \left(\mathrm{yx}\right)(1+{\tan }^2\left(\mathrm{yx}\right))$$$$\mathrm{we}\mathrm{have}\frac{\partial \mathrm{f}}{\partial \mathrm{x}}(\mathrm{x},\mathrm{y})=\frac{1}{\mathrm{x}}+\mathrm{y}\{1+{\tan }^2\left(\mathrm{xy}\right)\}\Rightarrow$$$$\frac{\partial }{\partial \mathrm{y}}\left(\frac{\partial \mathrm{f}}{\partial \mathrm{x}}(\mathrm{x},\mathrm{y})\right)=1+{\tan }^2\left(\mathrm{xy}\right)+\mathrm{y}2\mathrm{xtan}\left(\mathrm{xy}\right)(1+{\tan }^2\left(\mathrm{xy}\right))\Rightarrow$$$$\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{y}\partial \mathrm{x}}(\mathrm{x},\mathrm{y})=1+{\tan }^2\left(\mathrm{xy}\right)+2\mathrm{xytan}\left(\mathrm{xy}\right)(1+{\tan }^2\left(\mathrm{xy}\right))$$

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