If \(y = t -\displaystyle\frac{1}{t}\) and \(x = t +\displaystyle\frac{1}{t}\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{t^2+1}{t^2-1}\)
By parametric differentiation, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).
\( \displaystyle \frac{dy}{dt} = 1 + \frac{1}{t^2} \)
\( \displaystyle \frac{dx}{dt} = 1 - \frac{1}{t^2} \)
\( \displaystyle \frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \)
Multiplying the numerator and the denominator by \(t^2\):
\( \displaystyle \frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1} \)