Question

section 2.5: product and quotient rules (homework)
score: 0/80 answered: 0/8
question 1
if $f(t)=(t^{2}+3t + 2)(5t^{2}+6)$, find $f(t)$.
find $f(1)$.
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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 $u(t)=t^{2}+3t + 2$ and $v(t)=5t^{2}+6$.

Step2: Differentiate $u(t)$

Differentiate $u(t)=t^{2}+3t + 2$ using the power - rule. $u^\prime(t)=\frac{d}{dt}(t^{2}+3t + 2)=2t + 3$.

Step3: Differentiate $v(t)$

Differentiate $v(t)=5t^{2}+6$ using the power - rule. $v^\prime(t)=\frac{d}{dt}(5t^{2}+6)=10t$.

Step4: Apply product - rule

$f^\prime(t)=u^\prime(t)v(t)+u(t)v^\prime(t)=(2t + 3)(5t^{2}+6)+(t^{2}+3t + 2)\times10t$.
Expand the expressions:
\[

$$\begin{align*} (2t + 3)(5t^{2}+6)&=2t\times5t^{2}+2t\times6+3\times5t^{2}+3\times6\\ &=10t^{3}+12t + 15t^{2}+18\\ (t^{2}+3t + 2)\times10t&=10t^{3}+30t^{2}+20t \end{align*}$$

\]
Then $f^\prime(t)=10t^{3}+12t + 15t^{2}+18+10t^{3}+30t^{2}+20t=20t^{3}+45t^{2}+32t + 18$.

Step5: Find $f^\prime(1)$

Substitute $t = 1$ into $f^\prime(t)$.
$f^\prime(1)=20\times1^{3}+45\times1^{2}+32\times1+18=20 + 45+32+18=115$.

Answer:

$f^\prime(t)=20t^{3}+45t^{2}+32t + 18$
$f^\prime(1)=115$