If \[u={{\tan }^{-1}}(x+y),\] then \[x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\] [EAMCET 1996]

If \[u={{\tan }^{-1}}(x+y),\] then \[x\frac{\partial u}{\partial x}+y\

1.	If \[u={{\tan }^{-1}}(x+y),\] then \[x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\] [EAMCET 1996]
A.	\[\sin 2u\]
B.	\[\frac{1}{2}\sin 2u\]
C.	\[2\tan u\]
D.	\[{{\sec }^{2}}u\]
Answer» C. \[2\tan u\]

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If \[u={{\tan }^{-1}}(x+y),\] then \[x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\] [EAMCET 1996]

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