Let 
\[h\] be a vector-valued function defined by 
\[h(t)=(3t^2+t,2\cdot3^t)\].
Find 
\[h'(t)\].

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.

Learn how to find the derivative of the vector-valued function h(t) = (3t^2 + t, 2 * 3^t). Understand differentiation rules for polynomials and exponential functions. Suitable for undergraduate students studying calculus.

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