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

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(B) Given: $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$.Squaring the first equation on both sides,we get:$(x^2+y^2)^2 = (t+\frac{1}{t})^2$$

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$\frac{-y}{x}$

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(B) Given: $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$.Squaring the first equation on both sides,we get:$(x^2+y^2)^2 = (t+\frac{1}{t})^2$$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}+2$.Since $x^4+y^4=t^2+\frac{1}{t^2}$,we substitute this into the equation:$(t^2+\frac{1}{t^2})+2x^2y^2 = t^2+\frac{1}{t^2}+2$.This simplifies to $2x^2y^2 = 2$,which implies $x^2y^2 = 1$.Differentiating both sides with respect to $x$:$\frac{d}{dx}(x^2y^2) = \frac{d}{dx}(1)$$x^2(2y\frac{dy}{dx}) + y^2(2x) = 0$.$2x^2y\frac{dy}{dx} = -2xy^2$.$\frac{dy}{dx} = \frac{-2xy^2}{2x^2y} = \frac{-y}{x}$.

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

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