Answer
$\frac{dy}{dx} = \frac{e^t}{2t}$
$\frac{d^2y}{dx^2} = \frac{2e^t(t-1)}{8t^3}$
The curve is concave upward when $~~t \lt 0~~$ or $~~t \gt 1$
Work Step by Step
$x = t^2+1$
$y = e^t-1$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^t}{2t}$
We can find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac{2te^t-2e^t}{4t^2}}{2t} = \frac{2e^t(t-1)}{8t^3}$
The curve is concave upward when $\frac{d^2y}{dx^2} \gt 0$
$2e^t(t-1) \gt 0~~$ when $~~t \gt 1$
$8t^3 \gt 0~~$ when $~~t \gt 0$
The curve is concave upward when $~~t \gt 1$
$2e^t(t-1) \lt 0~~$ when $~~t \lt 1$
$8t^3 \lt 0~~$ when $~~t \lt 0$
The curve is concave upward when $~~t \lt 0$
The curve is concave upward when $~~t \lt 0~~$ or $~~t \gt 1$
