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

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

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

Given:

$y = t - \frac{1}{t}$

$x = t + \frac{1}{t}$

Compute derivatives:

$\frac{dy}{dt} = 1 + \frac{1}{t^{2}}$

$\frac{dx}{dt} = 1 - \frac{1}{t^{2}}$

Thus:

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

Multiply numerator and denominator by $t^{2}$:

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

$\displaystyle \frac{dy}{dx} = \frac{t^{2} + 1}{t^{2} - 1}$.