Answer
\[\frac{dy}{dx}=\frac{e^t}{2t}\]
\[\frac{d^2y}{dx^2}=\frac{e^t(t-1)}{4t^3}\]
\[t\in (-\infty,0)\cup (1,\infty)\]
Work Step by Step
\[x=t^2+1\;,\;y=e^{t}-1\]
Differentiate $x$ with respect to $t$:
\[\frac{dx}{dt}=2t\;\;\;\;\;\ldots (1)\]
Differentiate $y$ with respect to $t$:
\[\frac{dy}{dt}=e^t\]
\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{e^t}{2t}\]
\[\Rightarrow \frac{dy}{dx}=\frac{e^t}{2t}\]
Differeniate $\displaystyle\frac{dy}{dx}$ with respect to $x$ using chain rule:
\[\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}\]
\[\frac{d^2y}{dx^2}=\frac{d}{dt}\left[\frac{e^t}{2t}\right]\cdot\frac{dt}{dx}\]
\[\frac{d^2y}{dx^2}=\frac{1}{2}\left[\frac{e^tt-e^t(1)}{t^2}\right]\cdot\frac{dt}{dx}\]
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{1}{2}\left[\frac{e^t(t-1)}{t^2}\right]\cdot\frac{dt}{dx}\]
Using (1)
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{1}{2}\left[\frac{e^t(t-1)}{t^2}\right]\cdot \frac{1}{2t}\]
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{e^t(t-1)}{4t^3}\]
$\displaystyle \frac{d^2y}{dx^2}>0$ if $t<0$ and $t>1$ so The curve is concave up if $t\in (-\infty,0)\cup (1,\infty)$.
