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

For parametric equations where both $x$ and $y$ depend on parameter $t$:

$x = 4t$

$y = \frac{4}{t}$

The first derivative is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

---

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

$y = 4t^{-1}$

$\frac{dy}{dt} = -4t^{-2}$

$\frac{dy}{dt} = -\frac{4}{t^2}$

Therefore:

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

$\frac{dy}{dx} = -\frac{1}{t^2}$

---

For the second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx}$

$\frac{dy}{dx} = -t^{-2}$

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

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

---

Since $\frac{dx}{dt} = 4$:

$\frac{dt}{dx} = \frac{1}{4}$

Therefore:

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

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

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