If x^3+3xy+y^3=1,then the correct option(s) i | vedclass

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(C) Given equation: $x^3+3xy+y^3=1$Differentiating both sides with respect to $x$:$\frac{d}{dx}(x^3) + \frac{d}{dx}(3xy) + \frac{d}{dx}(y^3) = \frac{d

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$A$ and $C$

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(C) Given equation: $x^3+3xy+y^3=1$Differentiating both sides with respect to $x$:$\frac{d}{dx}(x^3) + \frac{d}{dx}(3xy) + \frac{d}{dx}(y^3) = \frac{d}{dx}(1)$$3x^2 + 3(y + x\frac{dy}{dx}) + 3y^2\frac{dy}{dx} = 0$Dividing by $3$: $x^2 + y + x\frac{dy}{dx} + y^2\frac{dy}{dx} = 0$$\frac{dy}{dx}(x + y^2) = -(x^2 + y)$$\frac{dy}{dx} = -\frac{x^2+y}{x+y^2}$At point $(1,1)$: $\frac{dy}{dx} = -\frac{1^2+1}{1+1^2} = -\frac{2}{2} = -1$. (Option $A$ is correct).At point $(1,0)$: $\frac{dy}{dx} = -\frac{1^2+0}{1+0^2} = -\frac{1}{1} = -1$. (Option $C$ is correct).Thus,options $A$ and $C$ are correct.

If $x^3+3xy+y^3=1$,then the correct option$(s)$ is/are:-
$(A) \left(\frac{dy}{dx}\right)_{(1,1)}=-1$
$(B) \left(\frac{dy}{dx}\right)_{(1,1)}=-2$
$(C) \left(\frac{dy}{dx}\right)_{(1,0)}=-1$
$(D) \left(\frac{dy}{dx}\right)_{(1,0)}=-3$

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If $x^3+3xy+y^3=1$,then the correct option$(s)$ is/are:-$(A) \left(\frac{dy}{dx}\right)_{(1,1)}=-1$$(B) \left(\frac{dy}{dx}\right)_{(1,1)}=-2$$(C) \left(\frac{dy}{dx}\right)_{(1,0)}=-1$$(D) \left(\frac{dy}{dx}\right)_{(1,0)}=-3$