Find frac{dy}{dx} for the equation x^{3}+x^{2 | vedclass

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(A) The given equation is $x^{3}+x^{2}y+xy^{2}+y^{3}=81$. Differentiating both sides with respect to $x$,we get: $\frac{d}{dx}(x^{3}+x^{2}y+xy^{2}+y^{

Vedclass · Accepted Answer

$\frac{-(3x^{2}+2xy+y^{2})}{x^{2}+2xy+3y^{2}}$

Vedclass · Answer

(A) The given equation is $x^{3}+x^{2}y+xy^{2}+y^{3}=81$. Differentiating both sides with respect to $x$,we get: $\frac{d}{dx}(x^{3}+x^{2}y+xy^{2}+y^{3}) = \frac{d}{dx}(81)$ Using the sum rule and product rule: $\frac{d}{dx}(x^{3}) + \frac{d}{dx}(x^{2}y) + \frac{d}{dx}(xy^{2}) + \frac{d}{dx}(y^{3}) = 0$ $3x^{2} + (x^{2}\frac{dy}{dx} + y(2x)) + (x(2y\frac{dy}{dx}) + y^{2}(1)) + 3y^{2}\frac{dy}{dx} = 0$ Grouping the terms containing $\frac{dy}{dx}$: $(x^{2} + 2xy + 3y^{2})\frac{dy}{dx} + (3x^{2} + 2xy + y^{2}) = 0$ Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-(3x^{2} + 2xy + y^{2})}{x^{2} + 2xy + 3y^{2}}$

Find $\frac{dy}{dx}$ for the equation $x^{3}+x^{2}y+xy^{2}+y^{3}=81$.

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