2.5 product and quotient rules\ndue thursday by 11:59pm points 3 submitting an external tool\n2.4\nscore…

Explanation:

Step1: Identify the product rule

The product rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=t^{2}+6t + 8$ and $v = 7t^{2}+2$.

Step2: Find $u'$ and $v'$

Differentiate $u$ with respect to $t$: $u'=\frac{d}{dt}(t^{2}+6t + 8)=2t+6$. Differentiate $v$ with respect to $t$: $v'=\frac{d}{dt}(7t^{2}+2)=14t$.

Step3: Apply the product - rule

$f'(t)=u'v+uv'=(2t + 6)(7t^{2}+2)+(t^{2}+6t + 8)\times14t$. Expand the first part: $(2t + 6)(7t^{2}+2)=14t^{3}+4t+42t^{2}+12$. Expand the second part: $(t^{2}+6t + 8)\times14t=14t^{3}+84t^{2}+112t$. Combine like - terms: $f'(t)=(14t^{3}+4t+42t^{2}+12)+(14t^{3}+84t^{2}+112t)=28t^{3}+126t^{2}+116t + 12$.

Step4: Find $f'(3)$

Substitute $t = 3$ into $f'(t)$: $f'(3)=28\times3^{3}+126\times3^{2}+116\times3+12$. $28\times3^{3}=28\times27 = 756$. $126\times3^{2}=126\times9 = 1134$. $116\times3=348$. $f'(3)=756+1134+348+12=2250$.

Answer:

$f'(t)=28t^{3}+126t^{2}+116t + 12$; $f'(3)=2250$