Find derivative of 4e^1/3 cos(t/2 - pi/4)

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.

Learn how to find the derivative of the function 4e^(1/3) cos(t/2 - pi/4) using the chain rule and trigonometric differentiation. This step-by-step solution explores advanced calculus concepts and is suitable for students in undergraduate calculus or advanced high school math.

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