A) \[\frac{1}{12}z\]
B) \[\frac{1}{4}z\]
C) \[\frac{1}{3}z\]
D) \[\frac{7}{12}z\]
Correct Answer: D
Solution :
Clearly z is homogeneous is x, y of order \[\frac{4}{3}-\frac{3}{4}=\frac{7}{12}\] \[\therefore \] By Euler?s Theorem \[x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{7}{12}z\].
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