if ( f(t)=(t^{2}+6t + 5)(2t^{2}+2) ), find ( f(t) ).\nuse the product rule to find this derivative. do not…

Explanation:

Step1: Recall the product rule

The product rule states that if (y = u(t)v(t)), then (y^\prime=u^\prime(t)v(t)+u(t)v^\prime(t)). Let (u(t)=t^{2}+6t + 5) and (v(t)=2t^{2}+2).

Step2: Find (u^\prime(t)) and (v^\prime(t))

Differentiate (u(t)): (u^\prime(t)=\frac{d}{dt}(t^{2}+6t + 5)=2t + 6). Differentiate (v(t)): (v^\prime(t)=\frac{d}{dt}(2t^{2}+2)=4t).

Step3: Apply the product rule

(f^\prime(t)=u^\prime(t)v(t)+u(t)v^\prime(t)=(2t + 6)(2t^{2}+2)+(t^{2}+6t + 5)(4t))

Step4: Find (f^\prime(3))

First, substitute (t = 3) into (u^\prime(t)), (v(t)), (u(t)) and (v^\prime(t)): (u^\prime(3)=2\times3+6=12), (v(3)=2\times3^{2}+2=2\times9 + 2=20), (u(3)=3^{2}+6\times3+5=9 + 18+5=32), (v^\prime(3)=4\times3 = 12). Then (f^\prime(3)=u^\prime(3)v(3)+u(3)v^\prime(3)=12\times20+32\times12) (=240+384) (=624)

Answer:

(f^\prime(t)=(2t + 6)(2t^{2}+2)+(t^{2}+6t + 5)(4t)); (f^\prime(3)=624)