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

x3 + x2y + xy2 + y3 = 81Differentiating 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 &there4; `3x^2 + x^2"dy"/"dx" + y xx 2x + x xx 2y"dy"/"dx" + y^2 xx 1 + 3y^2"dy"/"dx"` = 0 &there4; `3x^2 + x^2"dy"/"dx" + 2xy + 2xy"dy"/"dx" + y^2 + 3y^2"dy"/"dx"` = 0 &there4; `(x^2 + 2xy + 3y^2)"dy"/"dx" = -3x^2 - 2xy - y^2` &there4; `"dy"/"dx" = (-(3x^2 + 2xy + y^2))/(x^2 + 2xy + 3y^2)`.

Find dydxdydxIf x3 + x2y + xy2 + y3 = 81 - Mathematics and Statistics

Question

Find $\frac{dy}{dx}$ If x³ + x²y + xy² + y³ = 81

Sum

Solution

x³ + x²y + xy² + y³ = 81
Differentiating both sides w.r.t. x, we get
$3 x^{2} + x^{2} \frac{dy}{dx} + y \frac{d}{dx} (x^{2}) + x \frac{d}{dx} (y^{2}) + y^{2} \frac{d}{dx} (x) + 3 y^{2} \frac{dy}{dx}$ = 0

∴ $3 x^{2} + x^{2} \frac{dy}{dx} + y \times 2 x + x \times 2 y \frac{dy}{dx} + y^{2} \times 1 + 3 y^{2} \frac{dy}{dx}$ = 0

∴ $3 x^{2} + x^{2} \frac{dy}{dx} + 2 x y + 2 x y \frac{dy}{dx} + y^{2} + 3 y^{2} \frac{dy}{dx}$ = 0

∴ $(x^{2} + 2 x y + 3 y^{2}) \frac{dy}{dx} = - 3 x^{2} - 2 x y - y^{2}$

∴ $\frac{dy}{dx} = \frac{- (3 x^{2} + 2 x y + y^{2})}{x^{2} + 2 x y + 3 y^{2}}$ .

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Derivatives of Composite Functions - Chain Rule

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Chapter 1: Differentiation - Exercise 1.3 [Page 40]

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Balbharati Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board

Chapter 1 Differentiation
Exercise 1.3 | Q 3.04 | Page 40

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