Applications Of Derivatives Question 196
Question: If $ u(x,y)=y\log x+x,\log y, $ then $ u_{x}u_{y}-u_{x}\log x-u_{y}\log y+\log x\log y= $
[EAMCET 2003]
Options:
A) 0
B) -1
C) 1
D) 2
Show Answer
Answer:
Correct Answer: C
Solution:
$ u(x,y)=y\log x+x\log y $
$ u_{x}=\frac{y}{x}+\log y;\ u_{y}=\log x+\frac{x}{y} $
Now, $ u_{x}u_{y}-u_{x}\log x-u_{y}\log y+\log x\log y $
$ =( \frac{y}{x}+\log y )( \log x+\frac{x}{y} )-\frac{y}{x}\log x-\log x\log y-\log x\log y $
$ -\frac{x}{y}\log y+\log x\log y=1 $ .
BETA
