If x = t + frac{1}{t} and y = t - frac{1}{t}, | vedclass

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(B) Given $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = 1 - \frac{1}{t^2} = \frac{t^

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$-4t^3(t^2 - 1)^{-3}$

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(B) Given $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = 1 - \frac{1}{t^2} = \frac{t^2 - 1}{t^2}$$\frac{dy}{dt} = 1 + \frac{1}{t^2} = \frac{t^2 + 1}{t^2}$Now,find $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{(t^2 + 1)/t^2}{(t^2 - 1)/t^2} = \frac{t^2 + 1}{t^2 - 1}$Next,differentiate $\frac{dy}{dx}$ with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{dt}\left( \frac{t^2 + 1}{t^2 - 1} \right) \cdot \frac{dt}{dx}$Using the quotient rule for $\frac{d}{dt}\left( \frac{t^2 + 1}{t^2 - 1} \right)$:$= \frac{(2t)(t^2 - 1) - (t^2 + 1)(2t)}{(t^2 - 1)^2} = \frac{2t^3 - 2t - 2t^3 - 2t}{(t^2 - 1)^2} = \frac{-4t}{(t^2 - 1)^2}$Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{t^2}{t^2 - 1}$,we have:$\frac{d^2y}{dx^2} = \left( \frac{-4t}{(t^2 - 1)^2} \right) \cdot \left( \frac{t^2}{t^2 - 1} \right) = \frac{-4t^3}{(t^2 - 1)^3} = -4t^3(t^2 - 1)^{-3}$.

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

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