A) 0
B) 2u
C) 1/u
D) u
Correct Answer: A
Solution :
\[u={{\log }_{e}}({{x}^{2}}+{{y}^{2}})+{{\tan }^{-1}}\left( \frac{y}{x} \right)\] \[\frac{\partial u}{\partial x}=\frac{2x}{{{x}^{2}}+{{y}^{2}}}+\frac{1}{1+\frac{{{y}^{2}}}{{{x}^{2}}}}.\left( -\frac{y}{{{x}^{2}}} \right)\] \[=\frac{2x-y}{{{x}^{2}}+{{y}^{2}}}\] \[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\frac{({{x}^{2}}+{{y}^{2}}).2-(2x-y)2x}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\] \[=\frac{2{{y}^{2}}-2{{x}^{2}}+2xy}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\] \[\frac{\partial u}{\partial y}=\frac{2y}{{{x}^{2}}+{{y}^{2}}}+\frac{1}{1+\frac{{{y}^{2}}}{{{x}^{2}}}}.\frac{1}{x}=\frac{2y+x}{{{x}^{2}}+{{y}^{2}}}\] \[\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\frac{({{x}^{2}}+{{y}^{2}}).2-(2y+x)2y}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\]= \[\frac{2{{x}^{2}}-2{{y}^{2}}-2xy}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\] \[\therefore \] \[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=0\].
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