Differentiation Question 83
Question: If $ u=\log (x^{3}+y^{3}+z^{3}-3xyz) $ , then $ ( \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} ) $
$ (x+y+z) $ =
[EAMCET 1996]
Options:
A) 0
B) 1
C) 2
D) 3
Show Answer
Answer:
Correct Answer: D
Solution:
$ u=\log (x^{3}+y^{3}+z^{3}-3xyz) $
$ \therefore $ $ \frac{\partial u}{\partial x}=\frac{3x^{2}-3yz}{x^{3}+y^{3}+z^{3}-3xyz} $ ; $ \frac{\partial u}{\partial y}=\frac{3y^{2}-3zx}{x^{3}+y^{3}+z^{3}-3xyz} $
$ \frac{\partial u}{\partial z}=\frac{3z^{2}-3xy}{x^{3}+y^{3}+z^{3}-3xyz} $
$ \therefore $ $ \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} $ = $ \frac{3(x^{2}+y^{2}+z^{2}-xy-yz-zx)}{(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)} $
= $ \frac{3}{x+y+z} $ .
$ \therefore $ $ (x+y+z)( \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} )=3 $ .
BETA
