Find `(dy)/(dx)` in each of the following: `e^(x-y)=log(x/y)`

Explanation

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

Differntiating both sides w.r.t. `x`,

`e^(x-y) (1-dy/dx) = 1/(x/y)[(y-xdy/dx)/y^2]`

`=>e^(x-y) (1-dy/dx) = (y/x)[(y-xdy/dx)/y^2]`

`=>-e^(x-y)dy/dx +1/ydy/dx = -e^(x-y) +1/x`

`=>dy/dx((1-e^(x-y)y)/y) = (1-e^(x-y))/x`

`=>dy/dx = y/x ((1-e^(x-y)))/((1-e^(x-y)y)`