If `x^y=e^(x-y);` then `(dy)/(dx)` is

Options

A. e
B. 1
C. 0
D. `-1`

Answer

C

Explanation

Given have, `x^(y)=e^(x-y)`
taking log on both sides, we get
`ylogx=(x-y)loge=(x-y)` . . . (i)
when x=1, then `y(log1)=(1-y)impliesy=1`
On differentiating both sides,
`y((1)/(x))+"logx".(dy)/(dx)=1-(dy)/(dx)`
`implies(dy)/(dx)(logx+1)=1-(y)/(x)`
`(dy)/(dx)(logx+1)=(x-y)/(x)`
`implies(dy)/(dx)=((x-y))/(x(logx+1))`
when x=1, then
`((dy)/(dx))=(1-1)/(1(log1+1))=0`