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

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(D) Given: $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$.Squaring the first equation: $(x^2+y^2)^2 = (t+\frac{1}{t})^2$.$x^4+y^4+2x^2y^2 = t

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$-\frac{1}{x^3 y}$

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(D) Given: $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$.Squaring the first equation: $(x^2+y^2)^2 = (t+\frac{1}{t})^2$.$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}+2$.Substituting $x^4+y^4 = t^2+\frac{1}{t^2}$ into the equation:$(t^2+\frac{1}{t^2}) + 2x^2y^2 = t^2+\frac{1}{t^2}+2$.$2x^2y^2 = 2 \implies x^2y^2 = 1$.Thus,$y^2 = \frac{1}{x^2}$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx} = -2x^{-3} = -\frac{2}{x^3}$.$\frac{dy}{dx} = -\frac{1}{x^3y}$.

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

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