Question

product rule: problem 3 (1 point) let $f(t)=(t^{2}+7t + 4)(2t^{2}+4)$. find $f(t)$. $f(t)=square$ find $f(3)$. $f(3)=square$

Explanation:

Step1: Recall 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 $u(t)=t^{2}+7t + 4$ and $v(t)=2t^{2}+4$.

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

Differentiate $u(t)$: $u^\prime(t)=\frac{d}{dt}(t^{2}+7t + 4)=2t + 7$. Differentiate $v(t)$: $v^\prime(t)=\frac{d}{dt}(2t^{2}+4)=4t$.

Step3: Apply product - rule

$f^\prime(t)=u^\prime(t)v(t)+u(t)v^\prime(t)=(2t + 7)(2t^{2}+4)+(t^{2}+7t + 4)\times4t$.
Expand the first term: $(2t + 7)(2t^{2}+4)=4t^{3}+8t+14t^{2}+28$.
Expand the second term: $(t^{2}+7t + 4)\times4t = 4t^{3}+28t^{2}+16t$.
Combine like - terms: $f^\prime(t)=(4t^{3}+14t^{2}+8t + 28)+(4t^{3}+28t^{2}+16t)=8t^{3}+42t^{2}+24t + 28$.

Step4: Evaluate $f^\prime(3)$

Substitute $t = 3$ into $f^\prime(t)$:
$f^\prime(3)=8\times3^{3}+42\times3^{2}+24\times3 + 28$.
$8\times3^{3}=8\times27 = 216$.
$42\times3^{2}=42\times9 = 378$.
$24\times3=72$.
$f^\prime(3)=216+378+72 + 28=694$.

Answer:

$f^\prime(t)=8t^{3}+42t^{2}+24t + 28$
$f^\prime(3)=694$