If $x = 4t$ and $y =\frac{4}{t}$, then $\frac{d^2y}{dx^2}$ is

$\frac{1}{2t^3}$

The correct answer is Option (3) → $\frac{1}{2t^3}$

Given: $x = 4t$, $y = \frac{4}{t}$

Step 1: $\frac{dy}{dt} = \frac{d}{dt}\left(\frac{4}{t}\right) = -\frac{4}{t^2}$

Step 2: $\frac{dx}{dt} = \frac{d}{dt}(4t) = 4$

$\Rightarrow \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-\frac{4}{t^2}}{4} = -\frac{1}{t^2}$

Step 3: Differentiate again to get $\frac{d^2y}{dx^2}$

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

$\frac{d}{dt} \left( -\frac{1}{t^2} \right) = \frac{2}{t^3}$, and $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{4}$

$\Rightarrow \frac{d^2y}{dx^2} = \frac{2}{t^3} \cdot \frac{1}{4} = \frac{1}{2t^3}$