Let $x = t^2,y=t^3$. Then $\frac{d^2y}{dx^2}$ is equal to

$\frac{3}{4t}$

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

Given $x=t^2,\;y=t^3$.

$\frac{dy}{dt}=3t^2,\;\frac{dx}{dt}=2t$.

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3t^2}{2t}=\frac{3t}{2}$.

$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(\frac{dy}{dx}\Big)=\frac{\frac{d}{dt}\Big(\frac{dy}{dx}\Big)}{\frac{dx}{dt}}$.

$\frac{d}{dt}\Big(\frac{dy}{dx}\Big)=\frac{d}{dt}\Big(\frac{3t}{2}\Big)=\frac{3}{2}$.

Hence, $\frac{d^2y}{dx^2}=\frac{\frac{3}{2}}{2t}=\frac{3}{4t}$.

$\frac{d^2y}{dx^2}=\frac{3}{4t}$