Answer
$\frac{dy}{dx}=\frac{1-3t^{2}}{2t}$
and
$\frac{d^{2}y}{dx^{2}}=\frac{-1-3t^{2}}{4t^{3}}$
Work Step by Step
Given: $x=1+t^{2}$ and $y=t-t^{3}$
$\frac{dx}{dt}=2t$
$\frac{dy}{dt}=1-3t^{2}$
$\frac{dy}{dx}=\frac{{dy}/{dt}}{{dx}/{dt}}=\frac{1-3t^{2}}{2t}$
$\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}({dy}/{dx})}{dx/dt}$
$=\frac{\frac{d}{dt}(\frac{1-3t^{2}}{2t})}{2t}$
$=\frac{-\frac{1}{2}t^{-2}-\frac{3}{2}}{2t}\times \frac{2t^{2}}{2t^{2}}$
$=\frac{-1-3t^{2}}{4t^{3}}$
Hence, $\frac{dy}{dx}=\frac{1-3t^{2}}{2t}$
and
$\frac{d^{2}y}{dx^{2}}=\frac{-1-3t^{2}}{4t^{3}}$
