Question

attempt 1: 10 attempts remaining. if f(t) = (t⁴ + 5t + 2)(2t⁻⁴ + 6t⁻²), find f’(t). f’(t) = submit answer next item

Explanation:

Step1: Identify the product rule

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

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

Differentiate \( u(t) \) with respect to \( t \). Using the power rule \( \frac{d}{dt}(t^n)=nt^{n - 1} \), we get:
\( u'(t)=\frac{d}{dt}(t^4)+\frac{d}{dt}(5t)+\frac{d}{dt}(2)=4t^{3}+5+0 = 4t^{3}+5 \)

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

Differentiate \( v(t) \) with respect to \( t \). Using the power rule:
\( v'(t)=\frac{d}{dt}(2t^{-4})+\frac{d}{dt}(6t^{-2})=2\times(- 4)t^{-5}+6\times(-2)t^{-3}=-8t^{-5}-12t^{-3} \)

Step4: Apply the product rule

\( f'(t)=u'(t)v(t)+u(t)v'(t) \)
Substitute \( u(t),u'(t),v(t),v'(t) \):
\[

$$\begin{align*} f'(t)&=(4t^{3}+5)(2t^{-4}+6t^{-2})+(t^4 + 5t + 2)(-8t^{-5}-12t^{-3})\\ &=4t^{3}\times2t^{-4}+4t^{3}\times6t^{-2}+5\times2t^{-4}+5\times6t^{-2}+t^4\times(-8t^{-5})+t^4\times(-12t^{-3})+5t\times(-8t^{-5})+5t\times(-12t^{-3})+2\times(-8t^{-5})+2\times(-12t^{-3})\\ &=8t^{-1}+24t^{1}+10t^{-4}+30t^{-2}-8t^{-1}-12t^{1}-40t^{-4}-60t^{-2}-16t^{-5}-24t^{-3}\\ &=(8t^{-1}-8t^{-1})+(24t - 12t)+(10t^{-4}-40t^{-4})+(30t^{-2}-60t^{-2})-16t^{-5}-24t^{-3}\\ &=12t - 30t^{-4}-30t^{-2}-16t^{-5}-24t^{-3} \end{align*}$$

\]

Answer:

\( 12t-16t^{-5}-24t^{-3}-30t^{-4}-30t^{-2} \) (or can be written as \( 12t-\frac{16}{t^{5}}-\frac{24}{t^{3}}-\frac{30}{t^{4}}-\frac{30}{t^{2}} \))