If $x=2t^2+3, y=3t^2+6t+5, $ then the value of $\frac{d^2y}{dx^2}$ is :

$-\frac{3}{8t^3}$

The correct answer is Option (2) → $-\frac{3}{8t^3}$

Given:

$x = 2t^2 + 3, \quad y = 3t^2 + 6t + 5$

For parametric equations:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Differentiate with respect to $t$:

$\frac{dx}{dt} = 4t$

$\frac{dy}{dt} = 6t + 6$

$\frac{dy}{dx} = \frac{6t + 6}{4t} = \frac{3(t + 1)}{2t}$

Now differentiate again with respect to $t$:

$\frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \frac{3(t + 1)}{2t} \right) = \frac{3}{2} \frac{d}{dt} \left( 1 + \frac{1}{t} \right) = -\frac{3}{2t^2}$

Using:

$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}}$

$\frac{d^2y}{dx^2} = \frac{-\frac{3}{2t^2}}{4t} = -\frac{3}{8t^3}$