Question

Solve the following:

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")`    .....(i)

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

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

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

`"e"^"x". (1) + "e"^"y" "dy"/"dx" = "e"^("x + y"). ["d"/"dx" "x" + "d"/"dx" "y"]  ...("d"/"dx" "x" = 1)`

∴ `"e"^"x" + "e"^"y" "dy"/"dx" = "e"^("x + y") [1 + "dy"/"dx"]`

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

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

∴ `["e"^"y" − ("e"^"x" + "e"^"y")] "dy"/"dx" = ("e"^"x" + "e"^"y") − "e"^"x"  ...["From (i)"]`

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

∴ `(- "e"^"x") "dy"/"dx" = ("e"^"y")`

∴ `"dy"/"dx" = - ("e"^"y")/("e"^"x")`

∴ `"dy"/"dx" = - "e"^("y - x")`