Differentiation Question 94
Question: If $ u={{\tan }^{-1}}(x+y), $ then $ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}= $
[EAMCET 1996]
Options:
A) $ \sin 2u $
B) $ \frac{1}{2}\sin 2u $
C) $ 2\tan u $
D) $ {{\sec }^{2}}u $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \tan u=x+y=x.( 1+\frac{y}{x} ) $
$ \therefore $ $ \tan u $ is homogeneous in $ x,y $ of order 1.
$ \therefore $ $ x\frac{\partial }{\partial x}(\tan u)+y\frac{\partial }{\partial y}(\tan u)=\tan u $
$ \therefore $ $ x{{\sec }^{2}}u\frac{\partial u}{\partial x}+y{{\sec }^{2}}u\frac{\partial u}{\partial y}=\tan u $
$ \therefore $ $ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\tan u.{{\cos }^{2}}u=\sin u\cos u $ = $ \frac{1}{2}\sin 2u $ .
BETA
