Question

Find `"dy"/"dx"` if `e^(e^(x - y)) = x/y`

Answer 1

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

Taking log on both side

`log e^(e^((x - y))) = log  (x/y)`

∴ `e^((x - y)) log e = log x - log y`

∴ e^x–y = log x – log y ....[∵ log e = 1]
Differentiating both sides w.r.t. x, we get
`e^(x - y)."d"/"dx"(x - y) = (1)/x - (1)/y"dy"/"dx"`

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

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

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

`((1 - ye^(x - y))/y)"dy"/"dx" = (1 - xe^(x - y))/x`

∴ `"dy"/"dx" = (y(1 - xe^(x - y)))/((x(1 - ye^(x - y))`.