If `x^(y)=e^(x-y)`, then `(dy)/(dx)` is equal to

Options

A. `1/(1+logx)`
B. `(logx)/((1+logx)^(2))`
C. `x/((1+logx)^(2))`
D. `(logx)/(1+logx)`

Answer

B

Explanation

We have `x^(y)-x^(x-y)`………………..i
Taking log on both sides of Eq. i
we get
`y log x=(x-y)log_(e)e=(x-y)`
`impliesy=x/(1+logx)`
On differentiating we get
`(dy)/(dx)=(1+logx-x xx 1/x)/((1+logx)^(2))`
`=(1+logx-1)/((1+log x)^(2))=(logx)/((1+logx)^(2))`