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The function provided is:

\[
r(t) = 4 \cdot 2^{\sqrt{t}}
\]

To find the derivative \( r'(t) \), we will use both the product rule and the chain rule.

### Steps:

1. **Product Rule**: The function is in the form \( r(t) = 4 \cdot f(t) \), where \( f(t) = 2^{\sqrt{t}} \). Since 4 is a constant, its derivative will not change, and we can focus on differentiating \( 2^{\sqrt{t}} \).

2. **Differentiate \( f(t) = 2^{\sqrt{t}} \)**:
   - The general derivative of an exponential function \( a^x \) is \( \frac{d}{dx}(a^x) = a^x \ln(a) \).
   - Here, \( x = \sqrt{t} \), so we must apply the chain rule. The derivative of \( \sqrt{t} \) is \( \frac{1}{2\sqrt{t}} \).

Thus, the derivative of \( f(t) = 2^{\sqrt{t}} \) is:

\[
f'(t) = 2^{\sqrt{t}} \ln(2) \cdot \frac{1}{2\sqrt{t}}
\]

3. **Combine Everything**:
   Since \( r(t) = 4 \cdot 2^{\sqrt{t}} \), we multiply the derivative of \( 2^{\sqrt{t}} \) by 4:

\[
r'(t) = 4 \cdot 2^{\sqrt{t}} \ln(2) \cdot \frac{1}{2\sqrt{t}} = \frac{4 \ln(2)}{2\sqrt{t}} \cdot 2^{\sqrt{t}}
\]

Simplify:

\[
r'(t) = \frac{2 \ln(2) \cdot 2^{\sqrt{t}}}{\sqrt{t}}
\]

This is the derivative of the given function.

Let me know if you'd like further details or have any questions.

Here are some related questions you might find helpful:

1. How do you apply the chain rule in general?
2. What is the product rule and when is it used?
3. Can the derivative of \( 2^x \) be generalized for any function of \( x \)?
4. How can you further simplify expressions involving logarithms in derivatives?
5. What is the importance of constants like 4 in the differentiation process?

**Tip**: When differentiating exponential functions with complex exponents, always apply the chain rule carefully to each part.

Find the derivative of the function: r(t) = 4 * 2^(√t)

Learn how to find the derivative of the function r(t) = 4 * 2^(√t) using the product rule and chain rule. This problem involves key calculus concepts, including differentiation, exponential functions, and the application of both the product rule and chain rule. Suitable for students in Grades 11-12.

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