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Question
If y = `"x"^3 + 3"xy"^2 + 3"x"^2"y"` Find `"dy"/"dx"`
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Solution
y = `"x"^3 + 3"xy"^2 + 3"x"^2"y"`
Differentiating both sides w.r.t.x, we get
`"dy"/"dx" = "d"/"dx" ("x"^3) + 3"d"/"dx" ("xy"^2) + 3 "d"/"dx" ("x"^2"y")`
∴ `"dy"/"dx" = "3x"^2 + 3["x" * "d"/"dx" ("y"^2) + "y"^2 * "d"/"dx" ("x")] + 3 ["x"^2 * "dy"/"dx" + "y" * "d"/"dx" ("x"^2)]`
∴ `"dy"/"dx" = 3["x"^2 + "x" * "2y" "dy"/"dx" + "y"^2 (1) + "x"^2 "dy"/"dx" + "y"("2x")]`
∴ `"dy"/"dx" - 6"xy" "dy"/"dx" - 3"x"^2 "dy"/"dx" = 3 ("x"^2 + "y"^2 + 2"xy")`
∴ `"dy"/"dx" (1 - "6xy" - 3"x"^2) = 3("x"^2 + "y"^2 + 2"xy")`
∴ `"dy"/"dx" = (3("x"^2 + "y"^2 + "2xy"))/(1 - "6xy" - 3"x"^2)`
∴ `"dy"/"dx" = (-3("x"^2 + "y"^2 + "2xy"))/("6xy" + 3"x"^2 - 1)`
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