Differentiation Question 93
Question: If $ z^{2}=\frac{{x^{1/2}}+{y^{1/2}}}{{x^{1/3}}+{y^{1/3}}} $ then $ x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}= $
[EAMCET 1999]
Options:
A) $ \frac{z}{6} $
B) $ \frac{z}{3} $
C) $ \frac{z}{2} $
D) $ \frac{z}{12} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ z^{2} $ is homogeneous in x, y of degree $ \frac{1}{6} $
$ \therefore $ $ x\frac{\partial }{\partial x}(z^{2})+y\frac{\partial }{\partial y}(z^{2})=\frac{1}{6}(z^{2}) $
Therefore $ 2xz\frac{\partial z}{\partial x}+2yz\frac{\partial z}{\partial y}=\frac{1}{6}z^{2} $
Therefore $ x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{1}{12}z $ .
BETA
