If `x^y=e^(x-y),` show that `(dy)/(dx)=(logx)/({log(x e)}^2)`

Explanation

`x^y = e^(x-y)`

Taking log both sides,

`log(x^y) = log(e^(x-y))`

`=>ylogx = (x-y)loge`

As `log e = 1`,

`=>ylogx+y = x`

`=>y(1+logx) = x`

`y = x/(1+logx)`

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