Question

product rule: problem 3 (1 point) let $f(t)=(t^{2}+6t + 5)(3t^{2}+2)$. find $f(t)$. $f(t)=$ find $f(3)$. $f(3)=$ note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor

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 $u(t)=t^{2}+6t + 5$ and $v(t)=3t^{2}+2$.

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

Differentiate $u(t)$ with respect to $t$: $u^\prime(t)=\frac{d}{dt}(t^{2}+6t + 5)=2t + 6$. Differentiate $v(t)$ with respect to $t$: $v^\prime(t)=\frac{d}{dt}(3t^{2}+2)=6t$.

Step3: Apply the product - rule

$f^\prime(t)=u^\prime(t)v(t)+u(t)v^\prime(t)=(2t + 6)(3t^{2}+2)+(t^{2}+6t + 5)\times6t$.
Expand the expressions:
\[

$$\begin{align*} (2t + 6)(3t^{2}+2)&=2t\times3t^{2}+2t\times2+6\times3t^{2}+6\times2\\ &=6t^{3}+4t + 18t^{2}+12 \end{align*}$$

\]
\[

$$\begin{align*} (t^{2}+6t + 5)\times6t&=6t^{3}+36t^{2}+30t \end{align*}$$

\]
Then $f^\prime(t)=6t^{3}+4t + 18t^{2}+12+6t^{3}+36t^{2}+30t=12t^{3}+54t^{2}+34t + 12$.

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

Substitute $t = 3$ into $f^\prime(t)$:
\[

$$\begin{align*} f^\prime(3)&=12\times3^{3}+54\times3^{2}+34\times3 + 12\\ &=12\times27+54\times9+102 + 12\\ &=324+486+102 + 12\\ &=924 \end{align*}$$

\]

Answer:

$f^\prime(t)=12t^{3}+54t^{2}+34t + 12$
$f^\prime(3)=924$