If $x=6 t, y=4 t^3$, then $\frac{d^2 y}{d x^2}$ at t = 1 is:

$\frac{2}{3}$

The correct answer is Option (4) → $\frac{2}{3}$

$x=6t,\;\; y=4t^3$

$\frac{dx}{dt}=6,\;\; \frac{dy}{dt}=12t^2$

$\frac{dy}{dx}=\frac{12t^2}{6}=2t^2$

$\frac{d^2y}{dx^2}=\frac{d}{dt}(2t^2)\div \frac{dx}{dt}$

$= \frac{4t}{6}=\frac{2t}{3}$

$t=1 \Rightarrow \frac{d^2y}{dx^2}=\frac{2}{3}$

$\frac{d^2y}{dx^2}=\frac{2}{3}$ at $t=1$.