Question

hw11 differentiation rules ii ($3.3)
score: 1/8 answered: 1/8
question 2
if $f(t)=(t^{2}+5t + 3)(4t^{2}+6)$, find $f(t)$.
$f(t)=$

Explanation:

Step1: Apply product - rule

The product - rule states that if \(y = u(t)v(t)\), then \(y'=u'(t)v(t)+u(t)v'(t)\). Let \(u(t)=t^{2}+5t + 3\) and \(v(t)=4t^{2}+6\).

Step2: Find \(u'(t)\)

Differentiate \(u(t)=t^{2}+5t + 3\) term - by - term. Using the power rule \((x^{n})'=nx^{n - 1}\), we have \(u'(t)=2t + 5\).

Step3: Find \(v'(t)\)

Differentiate \(v(t)=4t^{2}+6\) term - by - term. Using the power rule, \(v'(t)=8t\).

Step4: Substitute into product - rule

\(f'(t)=(2t + 5)(4t^{2}+6)+(t^{2}+5t + 3)\times8t\)
Expand the two products:
\((2t + 5)(4t^{2}+6)=8t^{3}+12t+20t^{2}+30\)
\((t^{2}+5t + 3)\times8t=8t^{3}+40t^{2}+24t\)

Step5: Combine like terms

\(f'(t)=(8t^{3}+12t+20t^{2}+30)+(8t^{3}+40t^{2}+24t)\)
\(f'(t)=8t^{3}+8t^{3}+20t^{2}+40t^{2}+12t + 24t+30\)
\(f'(t)=16t^{3}+60t^{2}+36t + 30\)

Answer:

\(16t^{3}+60t^{2}+36t + 30\)