Answer
$y'=-\dfrac{3t^{2}+4t}{(t^{3}+2t^{2}-1)^{2}}$
Work Step by Step
$y=\dfrac{1}{t^{3}+2t^{2}-1}$
Differentiate using the quotient rule:
$y'=\dfrac{(t^{3}+2t^{2}-1)(1)'-(1)(t^{3}+2t^{2}-1)'}{(t^{3}+2t^{2}-1)^{2}}=...$
$...=\dfrac{(t^{3}+2t^{2}-1)(0)-(1)(3t^{2}+4t)}{(t^{3}+2t^{2}-1)^{2}}=...$
Evaluate the products and simplify:
$...=\dfrac{-3t^{2}-4t}{(t^{3}+2t^{2}-1)^{2}}=-\dfrac{3t^{2}+4t}{(t^{3}+2t^{2}-1)^{2}}$