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प्रश्न
Find `"dy"/"dx"`If x3 + x2y + xy2 + y3 = 81
उत्तर
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)`.
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