## Calculus (3rd Edition)

f'(t)=6$t^{2}$+2t-4
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)