## Calculus 8th Edition

$y'=\frac{t(-3t-4)}{(t^3+2t^2-1)^2}$
$y=\frac{1}{t^3+2t^2-1}$ Use the quotient rule. $y'=\frac{(t^3+2t^2-1)(0)-(1)(3t^2+4t)}{(t^3+2t^2-1)^2}$ Distribute and simplify. $y'=\frac{-3t^2-4t}{(t^3+2t^2-1)^2}$ Factor. $y'=\frac{t(-3t-4)}{(t^3+2t^2-1)^2}$