Question

hw11 differentiation rules ii (target c1, c2, c5; §3.3)
score: 1/8 answered: 1/8
question 2
if (f(t)=(t^{2}+4t + 6)(6t^{2}+5)), find (f(t)).
f(t)=
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Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=t^{2}+4t + 6$ and $v = 6t^{2}+5$.

Step2: Find $u'$

Differentiate $u=t^{2}+4t + 6$ with respect to $t$. Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $u'=\frac{d}{dt}(t^{2})+\frac{d}{dt}(4t)+\frac{d}{dt}(6)=2t + 4$.

Step3: Find $v'$

Differentiate $v = 6t^{2}+5$ with respect to $t$. Using the power rule, $v'=\frac{d}{dt}(6t^{2})+\frac{d}{dt}(5)=12t$.

Step4: Calculate $f'(t)$

Substitute $u$, $u'$, $v$, and $v'$ into the product - rule formula:
\[

$$\begin{align*} f'(t)&=u'v+uv'\\ &=(2t + 4)(6t^{2}+5)+(t^{2}+4t + 6)\times12t\\ &=(2t\times6t^{2}+2t\times5+4\times6t^{2}+4\times5)+(t^{2}\times12t+4t\times12t + 6\times12t)\\ &=(12t^{3}+10t + 24t^{2}+20)+(12t^{3}+48t^{2}+72t)\\ &=12t^{3}+10t + 24t^{2}+20+12t^{3}+48t^{2}+72t\\ &=24t^{3}+72t^{2}+82t + 20 \end{align*}$$

\]

Answer:

$24t^{3}+72t^{2}+82t + 20$