Question

attempt 1: 10 attempts remaining. differentiate ( y = e^{3t}(t^2 + 3t) ). answer: ( \frac{dy}{dt} = ) submit answer next item

Explanation:

Step1: Identify the product rule

We have \( y = e^{3t}(t^2 + 3t) \), so we use the product rule: if \( y = u \cdot v \), then \( \frac{dy}{dt} = u'v + uv' \), where \( u = e^{3t} \) and \( v = t^2 + 3t \).

Step2: Differentiate \( u = e^{3t} \)

Using the chain rule, \( u' = 3e^{3t} \) (since the derivative of \( e^{kt} \) is \( ke^{kt} \), here \( k = 3 \)).

Step3: Differentiate \( v = t^2 + 3t \)

Using the power rule, \( v' = 2t + 3 \).

Step4: Apply the product rule

\( \frac{dy}{dt} = u'v + uv' = 3e^{3t}(t^2 + 3t) + e^{3t}(2t + 3) \)

Step5: Factor out \( e^{3t} \)

\( \frac{dy}{dt} = e^{3t}(3(t^2 + 3t) + 2t + 3) = e^{3t}(3t^2 + 9t + 2t + 3) = e^{3t}(3t^2 + 11t + 3) \)

Answer:

\( \frac{dy}{dt} = e^{3t}(3t^2 + 11t + 3) \)