If \[u={{\log }_{e}}({{x}^{2}}+{{y}^{2}})+{{\tan }^{-1}}\left( \frac{y}{x} \right)\], then \[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\] [EAMCET 2000]

If \[u={{\log }_{e}}({{x}^{2}}+{{y}^{2}})+{{\tan }^{-1}}\left( \frac{y

1.	If \[u={{\log }_{e}}({{x}^{2}}+{{y}^{2}})+{{\tan }^{-1}}\left( \frac{y}{x} \right)\], then \[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\] [EAMCET 2000]
A.	0
B.	2u
C.	1/u
D.	u
Answer» B. 2u

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If \[u={{\log }_{e}}({{x}^{2}}+{{y}^{2}})+{{\tan }^{-1}}\left( \frac{y}{x} \right)\], then \[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\] [EAMCET 2000]

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