Answer
Tangent line: $\quad y=9x-1$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=2}=108$
Work Step by Step
$\displaystyle \frac{dx}{dt}=-(t+1)^{-2},\qquad \displaystyle \frac{dy}{dt}=-(t-1)^{-2}$
$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-(t-1)^{-2}}{-(t+1)^{-2}}=\frac{(t+1)^{2}}{(t-1)^{2}}$
The slope of the tangent line at $t=2$ is
$\left. \displaystyle \frac{dy}{dx} \right|_{t=\pi/2}=\frac{(2+1)^{2}}{(2-1)^{2}}=9$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=2}$,
$\displaystyle \frac{dy'}{dt}=\frac{d}{dt}[\frac{(t+1)^{2}}{(t-1)^{2}}]=\frac{2(t+1)(t-1)^{2}-2(t-1)(t+1)^{2}}{(t-1)^{4}}$
$=\displaystyle \frac{2(t+1)(t-1)[t-1-(t+1)]}{(t-1)^{4}}$
$=\displaystyle \frac{2(t+1)[-2]}{(t-1)^{3}}$
$=\displaystyle \frac{-4(t+1)}{(t-1)^{3}}$
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{\frac{-4(t+1)}{(t-1)^{3}}}{-(t+1)^{-2}}=\frac{4(t+1)^{3}}{(t-1)^{3}}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=2}=\frac{4(2+1)^{3}}{(2-1)^{3}}=4(27)=108$
The point $P$ on the curve corresponding to $t=2$ is
$P(\displaystyle \frac{1}{2+1},\ \frac{2}{2-1})=(\frac{1}{3},2)$
Use the point-slope equation for the tangent line.
$y-2=9\displaystyle \cdot(x-\frac{1}{3})$
$y=9x-3+2$
$y=9x-1$
