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If x3+y3=xyx^3 + y^3 = xy, then dydx\frac{dy}{dx} is equal to :

Solution

✅ Correct Option: 2

Differentiate x3+y3=xyx^3 + y^3 = xy implicitly: 3x2+3y2dydx=y+xdydx3x^2 + 3y^2 \frac{dy}{dx} = y + x\frac{dy}{dx}. So dydx(3y2−x)=y−3x2\frac{dy}{dx}(3y^2 - x) = y - 3x^2, giving dydx=y−3x23y2−x\frac{dy}{dx} = \frac{y - 3x^2}{3y^2 - x}.

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