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If x^y=e^(x-y), Prove that dy/dx=logx/(...

If `x^y=e^(x-y)`, Prove that `dy/dx=logx/(1+logx)^2`

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`"We have "x^(y)=e^(x-y)`
`"or "e^(ylog x)=e^(x-y)" "[becausex^(y)=e^(log x^(y))=e^(y log x)]`
`"or "ylog x = x-y`
`"or y=(x)/(1+log x)`
On differentiating both the sides w.r.t. x, we get
`(dy)/(dx)=((1+log x)xx1-x(0+(1)/(x)))/((1+ log x)^(2))=(log x)/((1+ log x )^(2))`
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