Differentiation Question 96
Question: If $ u=x^{2}{{\tan }^{-1}}\frac{y}{x}-y^{2}{{\tan }^{-1}}\frac{x}{y} $ , then $ \frac{{{\partial }^{2}}u}{\partial x\partial y}= $
[Tamilnadu (Engg.) 2002]
Options:
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}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ u=x^{2}{{\tan }^{-1}}\frac{y}{x}-y^{2}( \frac{\pi }{2}-{{\tan }^{-1}}\frac{y}{x} )=(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{2y^{2}}{x^{2}+y^{2}}=\frac{x^{2}-y^{2}}{x^{2}+y^{2}} $ .
BETA
