Question

If xy = e^(x – y), then show that `dy/dx = (y(x-1))/(x(y+1)) .`

Answer 1

We have , `xy = e^(x-y)`

Taking log on both sides,

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

⇒ `log x + log y= (x-y)log e`

⇒ `logx + log y = (x - y) xx 1`

⇒ `log x+ log y = x-y`

⇒ `d/dx (log x) + d/dx (log y ) = d/(dr) (x) - dy/dx`

⇒ `1/x + 1/y dy/dx = 1 -dx /dx`

⇒` ( 1 +1/y) dy/dx = 1 -1/x`

⇒ `(y+1/y)dy/dx = (x-1)/x`

⇒ `dy/dx = (y(x-1))/(x(y+1))`

Hence proved.