## Calculus 8th Edition

$\frac{4}{9}$
Given: $x=t^{3}+6t+1$ and $y=2t-t^{2}$ $\frac{dx}{dt}=3t^{2}+6$ $\frac{dy}{dt}=2-2t$ $\frac{dy}{dx}=\frac{{dy}/{dt}}{{dx}/{dt}}$ $=\frac{2-2t}{3t^{2}+6}$ Let $m$ be the slope of the tangent line to the given curve. $m=\frac{dy}{dx}|_{t=-1}=\frac{2-2t}{3t^{2}+6}|_{t=-1}$ $=\frac{2-2\times (-1)}{3(-1)^{2}+6}$ Hence, $m=\frac{4}{9}$