Differentiation Question 71
Question: If $ z=\frac{{{(x^{4}+y^{4})}^{1/3}}}{{{(x^{3}+y^{3})}^{1/4}}} $ , then $ x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}= $
Options:
A) $ \frac{1}{12}z $
B) $ \frac{1}{4}z $
C) $ \frac{1}{3}z $
D) $ \frac{7}{12}z $
Show Answer
Answer:
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 $ .
BETA
