If $x=2t^3+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) → $-\frac{3}{8t^3}$

$x=2t^3+3$

$\frac{dx}{dt}=6t^2$

$y=3t^2+6t+5$

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

$⇒\frac{dy}{dx}=\frac{6t+6}{6t^2}=\frac{t+1}{t^2}=\frac{1}{t}+\frac{1}{t^2}$

$⇒\frac{d^2y}{dx^2}=\left(-\frac{1}{t^2}-\frac{2}{t^3}\right)×\frac{dt}{dx}$

$=-\frac{t+2}{t^3}×\frac{1}{6t^2}$