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Question
Find `"dy"/"dx"` if, yex + xey = 1
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Solution
yex + xey = 1
Differentiating both sides w.r.t. x, we get
`"d"/"dx" ("ye"^"x") + "d"/"dx" ("xe"^"y") = 0`
∴ `"y" "d"/"dx" ("e"^"x") + "e"^"x" "dy"/"dx" +"x" "d"/"dx" ("e"^"y") + "e"^"y" "d"/"dx" ("x") = 0`
∴ `"y" "e"^"x" + ("e"^"x") "dy"/"dx" + "x"("e"^"y") "dy"/"dx" + "e"^"y"`
∴ `("e"^"x" + "x""e"^"y") "dy"/"dx" = - ("e"^"y" + "y" "e"^"x")`
∴ `"dy"/"dx" = (- ("e"^"y" + "y" "e"^"x"))/("e"^"x" + "x""e"^"y")`
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