If \[z=\frac{y}{x}\left[ \sin \frac{x}{y}+\cos \left( 1+\frac{y}{x} \right) \right]\], then \[x\frac{\partial z}{\partial x}=\] [EAMCET 2002]

\[2y\frac{\partial z}{\partial y}\]

\[y\frac{\partial z}{\partial y}\]

\[-y\frac{\partial z}{\partial y}\]

\[2y\frac{\partial z}{\partial x}\]

If \[z=\frac{y}{x}\left[ \sin \frac{x}{y}+\cos \left( 1+\frac{y}{x} \r

1.	If \[z=\frac{y}{x}\left[ \sin \frac{x}{y}+\cos \left( 1+\frac{y}{x} \right) \right]\], then \[x\frac{\partial z}{\partial x}=\] [EAMCET 2002]
A.	\[y\frac{\partial z}{\partial y}\]
B.	\[-y\frac{\partial z}{\partial y}\]
C.	\[2y\frac{\partial z}{\partial y}\]
D.	\[2y\frac{\partial z}{\partial x}\]
Answer» C. \[2y\frac{\partial z}{\partial y}\]

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If \[z=\frac{y}{x}\left[ \sin \frac{x}{y}+\cos \left( 1+\frac{y}{x} \right) \right]\], then \[x\frac{\partial z}{\partial x}=\] [EAMCET 2002]

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