To find the derivative of the vector-valued function $h(t) = (3t^2 + t, 2 \cdot 3^t)$, we need to differentiate each component of the function with respect to $t$.

The function $h(t)$ is given as: $h(t) = \left(3t^2 + t, 2 \cdot 3^t\right)$

Step 1: Differentiate the first component

The first component of $h(t)$ is $3t^2 + t$.

Differentiate with respect to $t$: $\frac{d}{dt} \left(3t^2 + t\right) = 6t + 1$

Step 2: Differentiate the second component

The second component of $h(t)$ is $2 \cdot 3^t$.

Differentiate with respect to $t$ using the chain rule: $\frac{d}{dt} \left(2 \cdot 3^t\right) = 2 \cdot \frac{d}{dt} \left(3^t\right)$ The derivative of $3^t$ is $3^t \ln(3)$, so: $\frac{d}{dt} \left(2 \cdot 3^t\right) = 2 \cdot 3^t \ln(3)$

Step 3: Combine the results

The derivative of $h(t)$, denoted by $h'(t)$, is the vector composed of the derivatives of the individual components: $h'(t) = \left(6t + 1, 2 \cdot 3^t \ln(3)\right)$

So the final answer is: $h'(t) = (6t + 1, 2 \cdot 3^t \ln(3))$

Would you like more details or have any questions?

Here are 5 related questions to further your understanding:

1. What is the derivative of $h(t) = (t^3 + 2t, e^{2t})$?
2. How would you find the second derivative $h''(t)$ for the given $h(t)$?
3. Can you find the derivative of the vector function $h(t) = (\sin(t), \cos(t))$?
4. What is the importance of the chain rule in differentiating vector-valued functions?
5. How would you integrate the vector function $h(t)$?

Tip: When differentiating functions involving exponentials with a base other than $e$, always remember to apply the chain rule and include the natural logarithm of the base.