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If x^y=e^(x-y), show that (dy)/(dx)=(log...

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

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`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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