If $y = t -\frac{1}{t}$ and $x = t +\frac{1}{t}$, then $\frac{dy}{dx}$ is equal to
Answer & explanation
Correct answer: option 2
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 $t^{2}$:
$\frac{t^{2} + 1}{t^{2} - 1}$
$\displaystyle \frac{dy}{dx} = \frac{t^{2} + 1}{t^{2} - 1}$.