If \(x = 2t + 1\) and \(y = t^2 - 3t\), then \(\displaystyle\frac{d^2y}{dx^2}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{2}\)
In parametric form:
\( \displaystyle \frac{dx}{dt} = 2, \quad \frac{dy}{dt} = 2t - 3 \)
\( \displaystyle \frac{dy}{dx} = \frac{2t - 3}{2} \)
The second derivative is the derivative of \(\frac{dy}{dx}\) with respect to x, so after differentiating with respect to t we divide again by \(\frac{dx}{dt}\):
\( \displaystyle \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{2t - 3}{2}\right)}{\frac{dx}{dt}} = \frac{1}{2} \)
Why the other options are incorrect
Option (2) 1 — Incorrect. This is \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\); it still has to be divided by \(\frac{dx}{dt} = 2\).
Option (3) 2 — Incorrect. This is \(\frac{d^2y}{dt^2}\), the second derivative with respect to t, not x.
Option (4) \(\displaystyle\frac{2t - 3}{2}\) — Incorrect. This is the first derivative \(\frac{dy}{dx}\).