Answer
f'(t)=6$t^{2}$+2t-4
Work Step by Step
The product rule states that if f(t)=h(t)g(t), then f'(t)=g'(t)h(t)+g(t)h'(t). We can find the derivative of the function f(t)=(2t+1)($t^{2}$-2) by setting g(t)=2t+1 and h(t)=$t^{2}$-2, and applying the product rule.
f'(t)=$\frac{d}{dt}$[2t+1]($t^{2}$-2)+(2t+1)$\frac{d}{dt}$[$t^{2}$-2]
$\frac{d}{dt}$[2t+1]=2, using the power rule
$\frac{d}{dt}$[$t^{2}$-2]=2t, using the power rule
Therefore,
f'(t)=(2)($t^{2}$-2)+(2t+1)(2t)
=(2$t^{2}$-4)+(4$t^{2}$+2t) (Simplify)
=6$t^{2}$+2t-4 (Simplify)
