Categories > differentiation > Partial differentiation > If $u={{\log }_{e}}({{x}^{2}}+{{y}^{2}})+{{\tan }^{-1}}\left( \frac{y}{x} \right)$, then $\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=$

differentiation

Question: If $u={{\log }_{e}}({{x}^{2}}+{{y}^{2}})+{{\tan }^{-1}}\left( \frac{y}{x} \right)$, then $\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=$


1) 0
2) 2u
3) 1/u
4) u
Solution: Explanation: No Explanation
Partial differentiation

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