#### Answer

$3t^{2}+2t-8$

#### Work Step by Step

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