If x^2+y^2=t+frac{2}{t} and x^4+y^4=t^2+frac{ | vedclass

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(B) Given equations are $x^2+y^2=t+\frac{2}{t}$ and $x^4+y^4=t^2+\frac{4}{t^2}$.Squaring the first equation: $(x^2+y^2)^2 = (t+\frac{2}{t})^2$.$x^4+y^

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-$2$

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(B) Given equations are $x^2+y^2=t+\frac{2}{t}$ and $x^4+y^4=t^2+\frac{4}{t^2}$.Squaring the first equation: $(x^2+y^2)^2 = (t+\frac{2}{t})^2$.$x^4+y^4+2x^2y^2 = t^2+\frac{4}{t^2}+4$.Substituting the value of $x^4+y^4$ from the second equation: $(t^2+\frac{4}{t^2}) + 2x^2y^2 = t^2+\frac{4}{t^2}+4$.This simplifies to $2x^2y^2 = 4$, which gives $x^2y^2 = 2$.Thus, $y^2 = \frac{2}{x^2}$.Differentiating both sides with respect to $x$: $2y \frac{dy}{dx} = -2 \cdot 2x^{-3} = -\frac{4}{x^3}$.Multiplying both sides by $x^3$: $2x^3y \frac{dy}{dx} = -4$.Therefore, $x^3y \frac{dy}{dx} = -2$.

If $x^2+y^2=t+\frac{2}{t}$ and $x^4+y^4=t^2+\frac{4}{t^2}$, then $x^3 y \frac{d y}{d x}$ equals

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