## Calculus (3rd Edition)

$3t^{2}+2t-8$
Using the product rule, we have $\frac{dy}{dt}=\frac{d}{dt}(t-8t^{-1})(t+t^{2})+(t-8t^{-1})\frac{d}{dt}(t+t^{2})$ $=(1+\frac{8}{t^{2}})(t+t^{2})+(t-\frac{8}{t})(1+2t)$ $=t+t^{2}+\frac{8}{t}+8+t+2t^{2}-\frac{8}{t}-16$ $=3t^{2}+2t-8$