Question

if ( f(t)=(t^{2}+6t + 5)(2t^{2}+2) ), find ( f(t) ).
use the product rule to find this derivative. do not simplify before or after applying the rule.

( f(t)= )

find ( f(3) ).

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:

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