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x^y = e^(x-y) so, prove that dy/dx = log...

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

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`x^y=e^(x-y)`
take `log_e`on both sides
`log_ex^y=log_e e(x-y)=> ylogx=(x-y)log_e e`
`ylox=x-y`..(1)
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