Differentiation Question 74
Question: If $ u=\frac{x+y}{x-y} $ , then $ \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}= $
[EAMCET 1991]
Options:
A) $ \frac{1}{x-y} $
B) $ \frac{2}{x-y} $
C) $ \frac{1}{{{(x-y)}^{2}}} $
D) $ \frac{2}{{{(x-y)}^{2}}} $
Show Answer
Answer:
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} $ .
BETA
