If \(x=\displaystyle\frac{1-t}{1+t}\) and \(y=\displaystyle\frac{2t}{1+t}\) then \(\displaystyle\frac{d^2y}{dx^2}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0
For a curve in parametric form, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). By the quotient rule:
\( \displaystyle \frac{dx}{dt} = \frac{-(1 + t) - (1 - t)}{(1 + t)^2} = \frac{-2}{(1 + t)^2} \)
\( \displaystyle \frac{dy}{dt} = \frac{2(1 + t) - 2t}{(1 + t)^2} = \frac{2}{(1 + t)^2} \)
\( \displaystyle \frac{dy}{dx} = \frac{2}{(1 + t)^2} \times \frac{(1 + t)^2}{-2} = -1 \)
dy/dx is a constant, so its derivative is zero:
\( \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}(-1) = 0 \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{1+t^2}\) — Incorrect. \(\frac{d^2y}{dx^2} = 0\) for every t.
Option (3) −1 — Incorrect. This is \(\frac{dy}{dx}\), the first derivative, not the second.
Option (4) \(\displaystyle\frac{-1}{1+t^2}\) — Incorrect. \(\frac{d^2y}{dx^2} = 0\) for every t.