Differentiation Question 81
Question: If $ u={{\tan }^{-1}}( \frac{x^{3}+y^{3}}{x-y} ) $ , then $ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}= $
[EAMCET 1999]
Options:
A) $ \sin 2u $
B) $ \cos 2u $
C) $ \tan 2u $
D) $ \sec 2u $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \tan u $ is homogeneous in x, y of degree 2.
$ \therefore $ $ x\frac{\partial }{\partial x}(\tan u)+y\frac{\partial }{\partial y}(\tan u)=2(\tan u) $
$ \therefore $ $ x{{\sec }^{2}}u\frac{\partial u}{\partial x}+y{{\sec }^{2}}u\frac{\partial u}{\partial y}=2\tan u $
Therefore $ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2\frac{\tan u}{{{\sec }^{2}}u} $ = $ 2\sin u\cos u=\sin 2u $ .
BETA
