If x^3 + y^3 = 6 and frac{d^2y}{dx^2} cdot fr | vedclass

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(A) Given $x^3 + y^3 = 6$. Differentiating with respect to $x$: $3x^2 + 3y^2 \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x^2}{y^2}$.Now, $\frac{

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$36$

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(A) Given $x^3 + y^3 = 6$. Differentiating with respect to $x$: $3x^2 + 3y^2 \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x^2}{y^2}$.Now, $\frac{d^2y}{dx^2} = -\frac{y^2(2x) - x^2(2y \frac{dy}{dx})}{y^4} = -\frac{2xy^2 - 2x^2y(-\frac{x^2}{y^2})}{y^4} = -\frac{2xy^3 + 2x^4}{y^5} = -\frac{2x(y^3 + x^3)}{y^5}$.Since $x^3 + y^3 = 6$, $\frac{d^2y}{dx^2} = -\frac{12x}{y^5}$.Similarly, by symmetry, $\frac{d^2x}{dy^2} = -\frac{12y}{x^5}$.Multiplying them: $\frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} = (-\frac{12x}{y^5})(-\frac{12y}{x^5}) = \frac{144}{x^4 y^4} = \frac{144}{(xy)^4}$.Comparing with $\frac{m}{(xy)^n}$, we get $m = 144$ and $n = 4$.Therefore, $\frac{m}{n} = \frac{144}{4} = 36$.

If $x^3 + y^3 = 6$ and $\frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} = \frac{m}{(xy)^n}$, where $m, n \in R$, then find the value of $\frac{m}{n}$.

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