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