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Question
Find `"dy"/"dx"` if, x3 + x2y + xy2 + y3 = 81
Solution
x3 + x2y + xy2 + y3 = 81
Differentiating both sides w.r.t. x, we get
`3"x"^2 + "x"^2 "dy"/"dx" + "y" * "d"/"dx" ("x"^2) + "x"*"d"/"dx" ("y"^2) + "y"^2 * "d"/"dx" ("x") + 3"y"^2 * "dy"/"dx" = 0`
∴ `3"x"^2 + "x"^2 "dy"/"dx" + "y" * "2x" + "x" * "2y" "dy"/"dx" + "y"^2 + 3"y"^2 * "dy"/"dx" = 0`
∴ `(3"x"^2 + 2"xy" + "y"^2) + ("x"^2 + 2"xy" + 3"y"^2) "dy"/"dx" = 0`
∴ `("x"^2 + 2"xy" + 3"y"^2) "dy"/"dx" = - (3"x"^2 + 2"xy" + "y"^2)`
∴ `"dy"/"dx" = - (3"x"^2 + 2"xy" + "y"^2)/("x"^2 + 2"xy" + 3"y"^2)`
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