If x = t², y = t³, then d²y/dx²

The correct option is 3/4t.

To find the second derivative of the parametric equations, we start with the given system of equations:
$x=t^2$
and
$y=t^3$

First, we compute the derivatives of $x$ and $y$ with respect to the parameter $t$:
$\frac{dx}{dt}=\frac{d}{dt}\left(t^2\right)=2t$
and
$\frac{dy}{dt}=\frac{d}{dt}\left(t^3\right)=3t^2$

Using the chain rule for parametric differentiation, the first derivative $\frac{dy}{dx}$ is given by:
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3t^2}{2t}=\frac{3}{2}t$

Next, we find the second derivative $\frac{d^{2}y}{d{x}^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$. We apply the chain rule once more:
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)·\frac{dt}{dx}$

We evaluate each part of the formula separately:
1. Differentiating $\frac{dy}{dx}$ with respect to $t$:
$\frac{d}{dt}\left(\frac{3}{2}t\right)=\frac{3}{2}$
2. Differentiating $t$ with respect to $x$ (which is the reciprocal of $\frac{dx}{dt}$):
$\frac{dt}{dx}=\frac{1}{dx/dt}=\frac{1}{2t}$

Finally, substituting these values back into the chain rule relation:
$\frac{d^{2}y}{d{x}^2}=\left(\frac{3}{2}\right)·\left(\frac{1}{2t}\right)=\frac{3}{4t}$