If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$, find the value of $\frac{dy}{dx}$.

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

The correct answer is Option (2) → $\frac{t^2 + 1}{t^2 - 1}$ ##

We have, $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$

Taking derivative of both equations w.r.t. $t$, we get

$∴\frac{dx}{dt} = \frac{d}{dt} \left( t + \frac{1}{t} \right) \quad \text{and} \quad \frac{dy}{dt} = \frac{d}{dt} \left( t - \frac{1}{t} \right)$

$\Rightarrow \frac{dx}{dt} = 1 + (-1)t^{-2} \quad \text{and} \quad \frac{dy}{dt} = 1 - (-1)t^{-2}$

$\Rightarrow \frac{dx}{dt} = 1 - \frac{1}{t^2} \quad \text{and} \quad \frac{dy}{dt} = 1 + \frac{1}{t^2}$

$\Rightarrow \frac{dx}{dt} = \frac{t^2 - 1}{t^2} \quad \text{and} \quad \frac{dy}{dt} = \frac{t^2 + 1}{t^2}$

$∴\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{t^2 + 1/t^2}{t^2 - 1/t^2} = \frac{t^2 + 1}{t^2 - 1}$