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) \[-\frac{{{x}^{2}}{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\]
Correct Answer: B
Solution :
\[u={{x}^{2}}{{\tan }^{-1}}\frac{y}{x}-{{y}^{2}}\left( \frac{\pi }{2}-{{\tan }^{-1}}\frac{y}{x} \right)=({{x}^{2}}+{{y}^{2}}){{\tan }^{-1}}\frac{y}{x}-\frac{\pi }{2}{{y}^{2}}\] \[\therefore \] \[\frac{\partial u}{\partial y}=({{x}^{2}}+{{y}^{2}})\frac{1}{1+\frac{{{y}^{2}}}{{{x}^{2}}}}.\frac{1}{x}+2y{{\tan }^{-1}}\frac{y}{x}-\pi y\] = \[x+2y{{\tan }^{-1}}\frac{y}{x}-\pi y\] \[\frac{{{\partial }^{2}}u}{\partial x\,\partial y}=1+2y\frac{1}{1+\frac{{{y}^{2}}}{{{x}^{2}}}}.\frac{-y}{{{x}^{2}}}\]=\[1-\frac{2{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}=\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\].
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