Question

Find `"dy"/"dx"`If x³ + x²y + xy² + y³ = 81

Answer 1

x³ + x²y + xy² + y³ = 81
Differentiating both sides w.r.t. x, we get
`3x^2 + x^2 "dy"/"dx" + y"d"/"dx"(x^2) + x"d"/"dx"(y^2) + y^2"d"/"dx"(x) + 3y^2"dy"/"dx"` = 0

∴ `3x^2 + x^2"dy"/"dx" + y xx 2x + x xx 2y"dy"/"dx" + y^2 xx 1 + 3y^2"dy"/"dx"` = 0

∴ `3x^2 + x^2"dy"/"dx" + 2xy + 2xy"dy"/"dx" + y^2 + 3y^2"dy"/"dx"` = 0

∴ `(x^2 + 2xy + 3y^2)"dy"/"dx" = -3x^2 - 2xy - y^2`

∴ `"dy"/"dx" = (-(3x^2 + 2xy + y^2))/(x^2 + 2xy + 3y^2)`.