Answer
$\frac{dy}{dx} = \frac{1-t}{e^{2t}}$
$\frac{d^2y}{dx^2} = \frac{2t-3}{e^{3t}}$
The curve is concave upward when $~~t \gt \frac{3}{2}$
Work Step by Step
$x = e^t$
$y = te^{-t}$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^{-t}-te^{-t}}{e^t} = \frac{1-t}{e^{2t}}$
We can find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac{-e^{2t}-2e^{2t}(1-t)}{e^{4t}}}{e^t} = \frac{2te^{2t}-3e^{2t}}{e^{5t}} = \frac{e^{2t}(2t-3)}{e^{5t}}= \frac{2t-3}{e^{3t}}$
The curve is concave upward when $\frac{d^2y}{dx^2} \gt 0$
$\frac{2t-3}{e^{3t}} \gt 0~~$ when $~~2t-3 \gt 0$
Then $~~t \gt \frac{3}{2}$
The curve is concave upward when $~~t \gt \frac{3}{2}$
