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To find the derivative of the function \( f(x) = \frac{2}{\sin(8x)} \), we can use the quotient rule and the chain rule. Here's the step-by-step process:

1. **Identify the function:** 
   The function can be rewritten as \( f(x) = 2 \cdot (\sin(8x))^{-1} \).

2. **Apply the derivative:**
   We use the chain rule where \( g(x) = \sin(8x) \) and \( h(x) = g(x)^{-1} \), so \( f(x) = 2 \cdot h(x) \). The derivative \( f'(x) \) is given by:
   \[
   f'(x) = 2 \cdot h'(x)
   \]
   To find \( h'(x) \), we have:
   \[
   h(x) = (\sin(8x))^{-1}
   \]
   Using the power rule and chain rule, we get:
   \[
   h'(x) = -1 \cdot (\sin(8x))^{-2} \cdot \cos(8x) \cdot 8
   \]
   Simplifying this, we have:
   \[
   h'(x) = -8 \cdot \frac{\cos(8x)}{\sin^2(8x)}
   \]

3. **Substitute back to find \( f'(x) \):**
   \[
   f'(x) = 2 \cdot (-8) \cdot \frac{\cos(8x)}{\sin^2(8x)}
   \]
   \[
   f'(x) = -16 \cdot \frac{\cos(8x)}{\sin^2(8x)}
   \]

Thus, the derivative of the function \( f(x) = \frac{2}{\sin(8x)} \) is:
\[
f'(x) = -16 \cdot \frac{\cos(8x)}{\sin^2(8x)}
\]

Would you like more details on this or have any questions about the steps? Here are some further questions you could consider:
1. How would the derivative change if the function was \( f(x) = \frac{2}{\cos(8x)} \)?
2. What are the critical points of \( f(x) = \frac{2}{\sin(8x)} \)?
3. How would you find the second derivative of this function?
4. How does the function behave near the points where \( \sin(8x) = 0 \)?
5. What applications might this function or its derivative have in real-world scenarios?

**Tip:** When dealing with trigonometric functions in derivatives, it can be helpful to sketch the functions to visualize their behavior and critical points.

Find the derivative of f(x) = 2 / sin(8x).

Learn how to compute the derivative of the function f(x) = 2 / sin(8x) using the quotient rule and chain rule. This detailed step-by-step solution explains key concepts in calculus, including trigonometric functions and differentiation. Suitable for high school and college calculus students.

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