A) \[\frac{1}{{{x}^{2}}+{{y}^{2}}}\]
B) 0
C) \[\frac{{{x}^{2}}-{{y}^{2}}}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\]
D) \[\frac{{{y}^{2}}-{{x}^{2}}}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\]
Correct Answer: B
Solution :
\[u=\log ({{x}^{2}}+{{y}^{2}})\,\therefore \frac{\partial u}{\partial x}=\frac{1}{{{x}^{2}}+{{y}^{2}}}.2x\] \[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\frac{({{x}^{2}}+{{y}^{2}}).2-2x.2x}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\]\[=\frac{2({{y}^{2}}-{{x}^{2}})}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\]? \[\frac{\partial u}{\partial y}=\frac{1}{{{x}^{2}}+{{y}^{2}}}.2y\] \[\therefore \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\frac{({{x}^{2}}+{{y}^{2}})\,.\,2-2y.2y}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}=\frac{2({{x}^{2}}-{{y}^{2}})}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\] \[\therefore \] \[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=0\].
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