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Question
Solve the following:
If `"e"^"x" + "e"^"y" = "e"^((x + y))` then show that, `"dy"/"dx" = - "e"^"y - x"`.
Solution
`"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")`
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