If $x=t^4, y=t$, then $\frac{d^2 y}{d x^2}$ is given by ________.

$-\frac{3}{16 t^7}$

The correct answer is Option (1) → $-\frac{3}{16 t^7}$

$x=t^4$ and $y=t$

$⇒\frac{dx}{dt}=4t^3$ and $⇒\frac{dy}{dt}=\frac{1}{4t^3}$

$⇒\frac{dy}{dx}=\frac{1}{4t^3}$

$⇒\frac{d^2y}{d x^2}=\frac{-3}{4t^3}×\frac{dt}{dx}$

$=\frac{-3}{4t^3}×\frac{1}{4t^3}$

$=-\frac{3}{16 t^3}$