If \(x = 4t\) and \(y = \displaystyle\frac{4}{t}\), then \(\displaystyle\frac{d^2y}{dx^2}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{2t^3}\)
For a curve in parametric form, the first derivative is:
\( \displaystyle \frac{dx}{dt} = 4,\qquad \frac{dy}{dt} = -\frac{4}{t^2} \)
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-4/t^2}{4} = -\frac{1}{t^2} \)
For the second derivative, differentiate \( \displaystyle\frac{dy}{dx} \) with respect to t and then divide by \( \displaystyle\frac{dx}{dt} \) again (chain rule):
\( \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dt}\left(-\frac{1}{t^2}\right) \cdot \frac{dt}{dx} \)
\( \displaystyle = \frac{2}{t^3} \cdot \frac{1}{4} = \frac{1}{2t^3} \)
Why the other options are incorrect
Option (1) \(-\displaystyle\frac{1}{t^2}\) — Incorrect. This is \( \displaystyle\frac{dy}{dx} \), the first derivative.
Option (2) \(\displaystyle\frac{2}{t^3}\) — Incorrect. This is only \( \displaystyle\frac{d}{dt}\left(\frac{dy}{dx}\right) \); it still has to be multiplied by \( \displaystyle\frac{dt}{dx} = \frac{1}{4} \).
Option (4) \(\displaystyle\frac{8}{t^3}\) — Incorrect. This multiplies \( \displaystyle\frac{2}{t^3} \) by \( \displaystyle\frac{dx}{dt} = 4 \) instead of dividing by it.