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The given expression is:

\[
V(t) = (t + 9e^{t}) \sqrt{t}
\]

To differentiate \( V(t) \), we will use the **product rule**, since this is a product of two functions: \( (t + 9e^{t}) \) and \( \sqrt{t} \).

### Steps:
1. **Identify the two functions**:
   \[
   u(t) = (t + 9e^{t}), \quad v(t) = \sqrt{t} = t^{1/2}
   \]
   
2. **Apply the product rule**:
   The product rule states:
   \[
   \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)
   \]

3. **Differentiate each function**:
   - \( u'(t) = \frac{d}{dt}(t + 9e^{t}) = 1 + 9e^{t} \)
   - \( v'(t) = \frac{d}{dt}(t^{1/2}) = \frac{1}{2}t^{-1/2} \)

4. **Substitute into the product rule**:
   \[
   V'(t) = (1 + 9e^{t}) \cdot \sqrt{t} + (t + 9e^{t}) \cdot \frac{1}{2}t^{-1/2}
   \]

5. **Simplify the expression**:
   \[
   V'(t) = (1 + 9e^{t}) \sqrt{t} + \frac{1}{2}(t + 9e^{t}) t^{-1/2}
   \]
   
   - First term: \( (1 + 9e^{t}) \sqrt{t} \)
   - Second term: \( \frac{1}{2}(t + 9e^{t}) t^{-1/2} \)

### Final differentiated result:

\[
V'(t) = (1 + 9e^{t}) \sqrt{t} + \frac{1}{2}(t + 9e^{t}) t^{-1/2}
\]

Would you like further clarification on any step?

Here are some related questions:
1. How do you apply the chain rule in similar problems?
2. Can the product rule be extended to three or more terms?
3. What are some common mistakes when differentiating exponential functions?
4. How do you handle the derivative of \( e^t \) in more complex expressions?
5. What other differentiation rules complement the product rule?

**Tip:** When differentiating functions involving products, always carefully apply the product rule first, and then simplify as needed.

Differentiate the function V(t) = (t + 9e^t)√t.

Learn how to differentiate the function V(t) = (t + 9e^t)√t using the product rule, exponential derivatives, and radical functions. A detailed, step-by-step solution is provided, focusing on key calculus concepts like the product rule and chain rule. Suitable for high school calculus students.

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