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.