If x^3 + y^3 - 3axy = 0,then frac{dy}{dx} equ | vedclass

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Question

(A) Given equation: $x^3 + y^3 - 3axy = 0$Differentiating both sides with respect to $x$:$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) - 3a \frac{d}{dx}(xy)

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given equation: $x^3 + y^3 - 3axy = 0$Differentiating both sides with respect to $x$:$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) - 3a \frac{d}{dx}(xy) = 0$Using the chain rule and product rule:$3x^2 + 3y^2 \frac{dy}{dx} - 3a \left( x \frac{dy}{dx} + y \right) = 0$Divide the entire equation by $3$:$x^2 + y^2 \frac{dy}{dx} - ax \frac{dy}{dx} - ay = 0$Group the terms containing $\frac{dy}{dx}$:$\frac{dy}{dx}(y^2 - ax) = ay - x^2$Therefore,$\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}$.

If $x^3 + y^3 - 3axy = 0$,then $\frac{dy}{dx}$ equals

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