To find the derivative of the function:

$f(t) = 4e^{1/3} \cos\left(\frac{t}{2} - \frac{\pi}{4}\right)$

First, notice that $e^{1/3}$ is a constant, and can be factored out of the derivative.

Step-by-Step Solution:

1. The function is of the form $f(t) = C \cos(g(t))$, where $C = 4e^{1/3}$ and $g(t) = \frac{t}{2} - \frac{\pi}{4}$.
2. The derivative of $\cos(g(t))$ is $-\sin(g(t)) \cdot g'(t)$, by the chain rule.

Finding $g'(t)$:

$g(t) = \frac{t}{2} - \frac{\pi}{4} \quad \Rightarrow \quad g'(t) = \frac{1}{2}$

Applying the derivative:

$f'(t) = C \cdot \left( -\sin\left( \frac{t}{2} - \frac{\pi}{4} \right) \right) \cdot g'(t)$ $f'(t) = 4e^{1/3} \cdot \left( -\sin\left( \frac{t}{2} - \frac{\pi}{4} \right) \right) \cdot \frac{1}{2}$

Simplifying:

$f'(t) = -2e^{1/3} \sin\left( \frac{t}{2} - \frac{\pi}{4} \right)$

Thus, the derivative of the function is:

$f'(t) = -2e^{1/3} \sin\left( \frac{t}{2} - \frac{\pi}{4} \right)$

Would you like further details on this process?

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Here are five related questions for further practice:

1. What is the derivative of $\sin\left(\frac{t}{2} + \frac{\pi}{4}\right)$?
2. How would the derivative change if $e^{1/3}$ was replaced by $e^{t/3}$?
3. Find the second derivative of $f(t) = 4e^{1/3} \cos\left(\frac{t}{2} - \frac{\pi}{4}\right)$.
4. How would the derivative change if $\cos$ was replaced with $\sin$?
5. What is the integral of $-2e^{1/3} \sin\left( \frac{t}{2} - \frac{\pi}{4} \right)$?

Tip: Always keep track of constants in derivative calculations to simplify the process.