Question

homework 3.3 differentiation rules
score: 70/120 answered: 7/12
question 8
textbook videos +
if ( f(t)=(t^{2}+7t + 2)(6t^{2}+6)), find ( f(t)).
find ( f(1)).
question help: video

Explanation:

Step1: Apply 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 $u(t)=t^{2}+7t + 2$ and $v(t)=5t^{2}+6$. First, find $u^\prime(t)$ and $v^\prime(t)$.
$u^\prime(t)=\frac{d}{dt}(t^{2}+7t + 2)=2t + 7$
$v^\prime(t)=\frac{d}{dt}(5t^{2}+6)=10t$

Step2: Calculate $f^\prime(t)$

Using the product - rule $f^\prime(t)=u^\prime(t)v(t)+u(t)v^\prime(t)$.
\[

$$\begin{align*} f^\prime(t)&=(2t + 7)(5t^{2}+6)+(t^{2}+7t + 2)\times10t\\ &=(2t\times5t^{2}+2t\times6+7\times5t^{2}+7\times6)+(t^{2}\times10t+7t\times10t + 2\times10t)\\ &=(10t^{3}+12t + 35t^{2}+42)+(10t^{3}+70t^{2}+20t)\\ &=10t^{3}+12t + 35t^{2}+42+10t^{3}+70t^{2}+20t\\ &=20t^{3}+105t^{2}+32t + 42 \end{align*}$$

\]

Step3: Find $f^\prime(1)$

Substitute $t = 1$ into $f^\prime(t)$.
$f^\prime(1)=20\times1^{3}+105\times1^{2}+32\times1+42$
$=20 + 105+32+42$
$=199$

Answer:

$f^\prime(t)=20t^{3}+105t^{2}+32t + 42$
$f^\prime(1)=199$