To find the derivative of the function \( y = \frac{1}{\sin(3x)} \), we can apply the chain rule and the derivative of the reciprocal function.

### Step-by-Step Solution:

1. **Express the function:**
   \[
   y = (\sin(3x))^{-1}
   \]

2. **Apply the chain rule:**
   The chain rule states that if \( y = (u(x))^{-1} \), then
   \[
   \frac{dy}{dx} = -u'(x) \cdot (u(x))^{-2}
   \]
   Here, \( u(x) = \sin(3x) \).

3. **Differentiate \( u(x) = \sin(3x) \):**
   The derivative of \( \sin(3x) \) with respect to \( x \) is
   \[
   u'(x) = \cos(3x) \cdot \frac{d}{dx}(3x) = 3\cos(3x)
   \]

4. **Apply the chain rule:**
   Substitute the derivative of \( u(x) \) into the chain rule formula:
   \[
   \frac{dy}{dx} = -3\cos(3x) \cdot (\sin(3x))^{-2}
   \]

5. **Simplify the expression:**
   \[
   \frac{dy}{dx} = -\frac{3\cos(3x)}{\sin^2(3x)}
   \]

So, the derivative of the function \( y = \frac{1}{\sin(3x)} \) is:

\[
\frac{dy}{dx} = -\frac{3\cos(3x)}{\sin^2(3x)}
\]

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

### Follow-up Questions:
1. How do you apply the chain rule to other trigonometric functions?
2. What is the derivative of \( y = \frac{1}{\cos(3x)} \)?
3. How does the quotient rule compare with the chain rule for this problem?
4. What happens if the function is \( y = \frac{1}{\sin^2(3x)} \)?
5. Can you find the second derivative of \( y = \frac{1}{\sin(3x)} \)?
6. How do you differentiate \( y = \frac{\sin(3x)}{\cos(3x)} \)?
7. What is the physical interpretation of this derivative if \( x \) represents time?
8. How do you integrate the derivative \( \frac{dy}{dx} = -\frac{3\cos(3x)}{\sin^2(3x)} \)?

### Tip:
Always verify the steps in differentiation, especially when applying multiple rules like the chain rule and product rule, to avoid mistakes.

Explore how to find the derivative of y = 1/sin(3x) using the chain rule and derivative of reciprocal function. This detailed guide covers trigonometric differentiation, suitable for advanced high school students and calculus enthusiasts.

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