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

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(B) Given equations are:$x^2+y^2=t+\frac{1}{t}$  ... $(i)$$x^4+y^4=t^2+\frac{1}{t^2}$  ... (ii)Squaring both sides of equation $(i)$,we get:$(x^2+y^2)

Vedclass · Accepted Answer

$-\frac{y}{x}$

Vedclass · Answer

(B) Given equations are:$x^2+y^2=t+\frac{1}{t}$  ... $(i)$$x^4+y^4=t^2+\frac{1}{t^2}$  ... (ii)Squaring both sides of equation $(i)$,we get:$(x^2+y^2)^2 = (t+\frac{1}{t})^2$$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}+2$Substituting the value from equation (ii) into this equation:$(x^4+y^4)+2x^2y^2 = (x^4+y^4)+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 $\frac{dy}{dx}$ is equal to

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