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

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(D) Given equations are:$x^4+y^4=t^2+\frac{1}{t^2} \quad \dots (i)$$x^2+y^2=t-\frac{1}{t} \quad \dots (ii)$Squaring equation $(ii)$ on both sides:$(x^

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

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(D) Given equations are:$x^4+y^4=t^2+\frac{1}{t^2} \quad \dots (i)$$x^2+y^2=t-\frac{1}{t} \quad \dots (ii)$Squaring equation $(ii)$ on both sides:$(x^2+y^2)^2 = (t-\frac{1}{t})^2$$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}-2$Substituting the value from equation $(i)$:$(t^2+\frac{1}{t^2}) + 2x^2y^2 = t^2+\frac{1}{t^2}-2$$2x^2y^2 = -2$$x^2y^2 = -1$$y^2 = -\frac{1}{x^2}$Differentiating both sides with respect to $x$:$\frac{d}{dx}(y^2) = \frac{d}{dx}(-x^{-2})$$2y \frac{dy}{dx} = -(-2)x^{-3}$$2y \frac{dy}{dx} = \frac{2}{x^3}$$\frac{dy}{dx} = \frac{1}{x^3y}$Thus,the correct option is $(d)$.

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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