Question

If e^x + e^y = e^{(x + y)}, then show that `dy/dx = -e^(y - x)`.

Answer 1

e^x + e^y = e^{x + y}                               ...[1]

Differentiating both sides w.r.t. x, we get

`e^x + e^y. dy/dx = e^(x + y).d/dx(x + y)`

∴ `e^x + e^y.dy/dx = e^(x + y).(1 + dy/dx)`

∴ `e^x + e^ydy/dx = e^(x + y) + e^(x + y)dy/dx`

∴ `(e^y - e^(x + y))dy/dx = e^(x + y)  –  e^x`

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

= `((e^x + e^y) - e^x)/(e^y - (e^x + e^y))`    ...[Using (1), since `e^(x + y) = e^x + e^y`]

= `e^y/-e^x`       

= `dy/dx − e^(y − x)`