Answer
Tangent line: $\quad y=-\displaystyle \frac{1}{2}x+\frac{1}{2}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=0}=-\frac{1}{8}$
Work Step by Step
$\displaystyle \frac{dx}{dt}=1+e^{t},\qquad \displaystyle \frac{dy}{dt}=-e^{t}$
$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-e^{t}}{1+e^{t}}$
The slope of the tangent line at $t=0$ is
$\left. \displaystyle \frac{dy}{dx} \right|_{t=0}=\dfrac{-1}{1+1}=-\dfrac{1}{2}$
The point $P$ on the curve corresponding to $t=0$ is
$P(1+1,\ 1-1)=(1,0)$
Use the point-slope equation for the tangent line.
$y-0=-\displaystyle \frac{1}{2}\cdot(x-1)$
$y=-\displaystyle \frac{1}{2}x+\frac{1}{2}$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=0}$,
$\displaystyle \frac{dy'}{dt}=\frac{d}{dt}[\frac{-e^{t}}{1+e^{t}}]=-\frac{e^{t}(1+e^{t})-e^{t}(1+e^{t})}{(1+e^{t})^{2}}$
$=-\displaystyle \frac{e^{t}(1+e^{t})-e^{t}(e^{t})}{(1+e^{t})^{2}}$
$=-\displaystyle \frac{e^{t}(1+e^{t}-e^{t})}{(1+e^{t})^{2}}$
$=-\displaystyle \frac{e^{t}}{(1+e^{t})^{2}}$
$ \displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{-\frac{e^{t}}{(1+e^{t})^{2}}}{1+e^{t}}=-\frac{e^{t}}{(1+e^{t})^{3}}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=0}=-\frac{1}{(1+1)^{3}}=-\frac{1}{8}$
