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

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(A) Given that $x^4+y^4=t^2+\frac{1}{t^2}$.We know that $(x^2+y^2)^2 = x^4+y^4+2x^2y^2$.Substituting the given values:$(t+\frac{1}{t})^2 = (t^2+\frac{

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

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

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