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

$\frac{1}{48}$

The correct answer is Option (4) → $\frac{1}{48}$

$x=6t^4$,    $y=4t^3$

$⇒\frac{dx}{dt}=24t^3$,     $\frac{dy}{dt}=12t^2$

$⇒\frac{dy}{dx}=\frac{1}{2t}$

$⇒\frac{d^2y}{dx^2}=\frac{1}{2t}×\frac{dt}{dx}=\frac{1}{48t^4}$

$\left.\frac{d^2y}{dx^2}\right|_{t=1}=\frac{1}{48}$