If $x=log\, t^2 $ and $y=(log\, t)^2 $ then $\frac{d^2y}{dx^2}$ is :

$\frac{1}{2}$

The correct answer is Option (2) → $\frac{1}{2}$

$x=\log t^2$

$x=2\log t$

$y=(\log t)^2$

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

$\frac{dy}{dt}=2\frac{(\log t)}{t}⇒\frac{dy}{dx}=\log t$

so $\frac{d^2y}{dx^2}=\frac{1}{t}\frac{dt}{dx}=\frac{1}{t}×\frac{t}{2}$

$⇒\frac{d^2y}{dx^2}=\frac{1}{2}$