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Let's find the derivative of the given function \( s = \frac{1}{4\pi} \sin(4t) - \frac{1}{3\pi} \cos(3t) \).

### Step-by-step process:

1. **Differentiate the first term:**
   \[
   \frac{d}{dt} \left( \frac{1}{4\pi} \sin(4t) \right)
   \]
   Using the chain rule:
   \[
   \frac{1}{4\pi} \cdot \frac{d}{dt} (\sin(4t)) = \frac{1}{4\pi} \cdot 4 \cos(4t) = \frac{1}{\pi} \cos(4t)
   \]

2. **Differentiate the second term:**
   \[
   \frac{d}{dt} \left( -\frac{1}{3\pi} \cos(3t) \right)
   \]
   Again, using the chain rule:
   \[
   -\frac{1}{3\pi} \cdot \frac{d}{dt} (\cos(3t)) = -\frac{1}{3\pi} \cdot (-3 \sin(3t)) = \frac{1}{\pi} \sin(3t)
   \]

3. **Combine the results:**
   \[
   \frac{ds}{dt} = \frac{1}{\pi} \cos(4t) + \frac{1}{\pi} \sin(3t)
   \]

### Final answer:
\[
\frac{ds}{dt} = \frac{1}{\pi} \cos(4t) + \frac{1}{\pi} \sin(3t)
\]

Would you like more details on any part of this solution?

Here are some related questions:
1. What is the chain rule, and why is it used here?
2. How do you find the derivative of sine and cosine functions?
3. How would the answer change if the angles inside the trigonometric functions were different?
4. Can you explain more about the steps involving constants during differentiation?
5. How do you apply the product or quotient rule in derivatives?

**Tip:** When differentiating trigonometric functions, always remember to apply the chain rule if the argument inside the function is more complex than just the variable (like \(4t\) or \(3t\) here).

Find the derivative of the function s = (1/(4π))sin(4t) - (1/(3π))cos(3t).

Find the exact derivative of the function s = (1/(4π))sin(4t) - (1/(3π))cos(3t) using the chain rule. This solution involves differentiation of trigonometric functions and constant multiples. Perfect for high school and introductory college calculus students.

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