If {x^3} + 8xy + {y^3} = 64,then frac{dy}{dx} | vedclass

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Question

(A) Given the equation: ${x^3} + 8xy + {y^3} = 64$Differentiating both sides with respect to $x$:$\frac{d}{dx}(x^3) + \frac{d}{dx}(8xy) + \frac{d}{dx}

Vedclass · Accepted Answer

$-\frac{3x^2 + 8y}{8x + 3y^2}$

Vedclass · Answer

(A) Given the equation: ${x^3} + 8xy + {y^3} = 64$Differentiating both sides with respect to $x$:$\frac{d}{dx}(x^3) + \frac{d}{dx}(8xy) + \frac{d}{dx}(y^3) = \frac{d}{dx}(64)$Applying the chain rule and product rule:$3x^2 + 8(y + x\frac{dy}{dx}) + 3y^2\frac{dy}{dx} = 0$Expanding the terms:$3x^2 + 8y + 8x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$Grouping the terms containing $\frac{dy}{dx}$:$(8x + 3y^2)\frac{dy}{dx} = -(3x^2 + 8y)$Solving for $\frac{dy}{dx}$:$\frac{dy}{dx} = -\frac{3x^2 + 8y}{8x + 3y^2}$Thus,the correct option is $A$.

If ${x^3} + 8xy + {y^3} = 64$,then $\frac{dy}{dx} = $

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