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,differentiate $x$ with respect to $t$:$\frac{dx}{dt} = \frac{d}{dt}(t + t^{-1}) = 1 -

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$\frac{t^2 + 1}{t^2 - 1}$

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(B) Given $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$.First,differentiate $x$ with respect to $t$:$\frac{dx}{dt} = \frac{d}{dt}(t + t^{-1}) = 1 - t^{-2} = 1 - \frac{1}{t^2} = \frac{t^2 - 1}{t^2}$.Next,differentiate $y$ with respect to $t$:$\frac{dy}{dt} = \frac{d}{dt}(t - t^{-1}) = 1 - (-t^{-2}) = 1 + \frac{1}{t^2} = \frac{t^2 + 1}{t^2}$.Now,use the chain rule for parametric differentiation:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{(t^2 + 1)/t^2}{(t^2 - 1)/t^2} = \frac{t^2 + 1}{t^2 - 1}$.

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{dy}{dx} = $

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