x^3+y^3=3xy Rightarrow frac{dy}{dx}= | vedclass

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Question

(A) Given the equation: $x^3+y^3=3xy$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(3xy)$. $3

Vedclass · Accepted Answer

$\frac{y-x^2}{y^2-x}$

Vedclass · Answer

(A) Given the equation: $x^3+y^3=3xy$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(3xy)$. $3x^2 + 3y^2 \frac{dy}{dx} = 3(y + x \frac{dy}{dx})$. Dividing by $3$: $x^2 + y^2 \frac{dy}{dx} = y + x \frac{dy}{dx}$. Rearranging the terms to isolate $\frac{dy}{dx}$: $y^2 \frac{dy}{dx} - x \frac{dy}{dx} = y - x^2$. $\frac{dy}{dx}(y^2 - x) = y - x^2$. Therefore,$\frac{dy}{dx} = \frac{y-x^2}{y^2-x}$.

$x^3+y^3=3xy \Rightarrow \frac{dy}{dx}=$

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