A) \[\sin 2u\]
B) \[\frac{1}{2}\sin 2u\]
C) \[2\tan u\]
D) \[{{\sec }^{2}}u\]
Correct Answer: B
Solution :
\[\tan u=x+y=x.\left( 1+\frac{y}{x} \right)\] \[\therefore \] \[\tan u\] is homogeneous in \[x,\,y\] of order 1. \[\therefore \] \[x\frac{\partial }{\partial x}(\tan u)+y\frac{\partial }{\partial y}(\tan u)=\tan u\] \[\therefore \] \[x{{\sec }^{2}}u\frac{\partial u}{\partial x}+y{{\sec }^{2}}u\frac{\partial u}{\partial y}=\tan u\] \[\therefore \] \[x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\tan u.{{\cos }^{2}}u=\sin u\cos u\] = \[\frac{1}{2}\sin 2u\].
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