Answer
$y'=\frac{t(-3t-4)}{(t^3+2t^2-1)^2}$
Work Step by Step
$y=\frac{1}{t^3+2t^2-1}$
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.
$y'=\frac{t(-3t-4)}{(t^3+2t^2-1)^2}$
