It looks like you're comparing the derivative you need to find with the expression given in your options. Let's revisit the derivative and rewrite it to match the options provided.

Given Function:

$y = \frac{1}{\sin(3x)}$

Rewriting $y$ using cosecant:

$y = \csc(3x)$

Step-by-Step Differentiation:

1. Differentiate using the chain rule:

   The derivative of $\csc(3x)$ is: $\frac{dy}{dx} = -\csc(3x) \cdot \cot(3x) \cdot \frac{d}{dx}(3x)$ Since $\frac{d}{dx}(3x) = 3$, we have: $\frac{dy}{dx} = -3\csc(3x) \cdot \cot(3x)$

2. Express $\cot(3x)$ in terms of $\cos(3x)$ and $\sin(3x)$: $\cot(3x) = \frac{\cos(3x)}{\sin(3x)}$

   Substituting into the derivative: $\frac{dy}{dx} = -3\csc(3x) \cdot \frac{\cos(3x)}{\sin(3x)}$

3. Simplify using the identity $\csc(3x) = \frac{1}{\sin(3x)}$: $\frac{dy}{dx} = -3 \cdot \frac{1}{\sin(3x)} \cdot \frac{\cos(3x)}{\sin(3x)}$

4. Final simplification: $\frac{dy}{dx} = -\frac{3\cos(3x)}{\sin^2(3x)}$

This expression can be rewritten as: $\frac{dy}{dx} = -3\cos(3x) \csc^2(3x)$

The expression you provided, $3 \csc^2(3x) \cos(3x)$, matches the derivative except for the sign. If you're looking for the positive version (without the negative sign), it might be listed that way in the options.

Would you like to clarify anything else?

Follow-up Questions:

1. Why does the negative sign appear in the original derivative?
2. How does $\csc(x)$ differ from $\sec(x)$ in differentiation?
3. What are the key identities to remember when differentiating trigonometric functions?
4. Can you find the derivative of $y = \cot(3x)$?
5. How does the chain rule affect the result when differentiating trigonometric functions?
6. What happens if $y = \sin^{-1}(3x)$?
7. How do trigonometric identities simplify the differentiation process?
8. How does differentiation of trigonometric functions apply in physics?

Tip:

When comparing your result with given options, consider possible sign changes and trigonometric identities that may alter the appearance of the expression without changing its mathematical meaning.