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

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(D) Given: $x^2+y^2=t-\frac{1}{t}$ (Equation $1$) Squaring both sides of Equation $1$: $(x^2+y^2)^2 = (t-\frac{1}{t})^2$ $x^4+y^4+2x^2y^2 = t^2+\frac{

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$\frac{-y}{x}$

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(D) Given: $x^2+y^2=t-\frac{1}{t}$ (Equation $1$) Squaring both sides of Equation $1$: $(x^2+y^2)^2 = (t-\frac{1}{t})^2$ $x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}-2$ Given: $x^4+y^4=t^2+\frac{1}{t^2}$ (Equation $2$) Substituting Equation $2$ into the expanded form: $(t^2+\frac{1}{t^2})+2x^2y^2 = t^2+\frac{1}{t^2}-2$ $2x^2y^2 = -2$ $x^2y^2 = -1$ 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 $\frac{dy}{dx}=$

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