If x^2+y^2=t+frac{1}{t} and x^4+y^4=t^2+frac{ | vedclass

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(A) Given, $x^4+y^4=t^2+\frac{1}{t^2}$.We know that $(t+\frac{1}{t})^2 = t^2 + \frac{1}{t^2} + 2$.Thus, $t^2 + \frac{1}{t^2} = (t+\frac{1}{t})^2 - 2$.

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(A) Given, $x^4+y^4=t^2+\frac{1}{t^2}$.We know that $(t+\frac{1}{t})^2 = t^2 + \frac{1}{t^2} + 2$.Thus, $t^2 + \frac{1}{t^2} = (t+\frac{1}{t})^2 - 2$.Substituting the given values, we get $x^4+y^4 = (x^2+y^2)^2 - 2$.Expanding the right side: $x^4+y^4 = x^4 + y^4 + 2x^2y^2 - 2$.Subtracting $x^4+y^4$ from both sides, we get $0 = 2x^2y^2 - 2$, which implies $x^2y^2 = 1$.Therefore, $y^2 = \frac{1}{x^2} = x^{-2}$.Differentiating both sides with respect to $x$: $\frac{d}{dx}(y^2) = \frac{d}{dx}(x^{-2})$.$2y \frac{dy}{dx} = -2x^{-3} = -\frac{2}{x^3}$.Dividing by $2$, we get $y \frac{dy}{dx} = -\frac{1}{x^3}$.Multiplying both sides by $x^3$, we get $x^3 y \frac{dy}{dx} = -1$.

If $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$, then $x^3 y \frac{d y}{d x}$ is equal to

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