If \[u={{x}^{2}}{{\tan }^{-1}}\frac{y}{x}-{{y}^{2}}{{\tan }^{-1}}\frac{x}{y}\], then \[\frac{{{\partial }^{2}}u}{\partial x\,\partial \,y}=\] [Tamilnadu (Engg.) 2002]

\[\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}\]

\[\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\]

\[-\frac{{{x}^{2}}{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\]

If \[u={{x}^{2}}{{\tan }^{-1}}\frac{y}{x}-{{y}^{2}}{{\tan }^{-1}}\frac

1.	If \[u={{x}^{2}}{{\tan }^{-1}}\frac{y}{x}-{{y}^{2}}{{\tan }^{-1}}\frac{x}{y}\], then \[\frac{{{\partial }^{2}}u}{\partial x\,\partial \,y}=\] [Tamilnadu (Engg.) 2002]
A.	\[\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}\]
B.	\[\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\]
C.	\[\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}\]
D.	\[-\frac{{{x}^{2}}{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\]
Answer» C. \[\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}\]

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If \[u={{x}^{2}}{{\tan }^{-1}}\frac{y}{x}-{{y}^{2}}{{\tan }^{-1}}\frac{x}{y}\], then \[\frac{{{\partial }^{2}}u}{\partial x\,\partial \,y}=\] [Tamilnadu (Engg.) 2002]

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