To find the derivative of \( f(x) = -\cos(x) \csc(x) \) with respect to \( x \), we can use the product rule and the derivatives of trigonometric functions.

### Step-by-Step Solution:

1. **Rewrite the Function:**
   \[
   f(x) = -\cos(x) \csc(x)
   \]
   Let \( u = -\cos(x) \) and \( v = \csc(x) \).

2. **Apply the Product Rule:**  
   The product rule states that \( \frac{d}{dx}(u \cdot v) = u'v + uv' \). We’ll need \( u' \) and \( v' \) here.

3. **Differentiate \( u = -\cos(x) \):**
   \[
   u' = \sin(x)
   \]

4. **Differentiate \( v = \csc(x) \):**
   \[
   v' = -\csc(x) \cot(x)
   \]

5. **Apply the Product Rule:**
   \[
   \frac{d}{dx}(-\cos(x) \csc(x)) = (\sin(x)) \csc(x) + (-\cos(x)) (-\csc(x) \cot(x))
   \]

6. **Simplify Each Term:**
   - The first term simplifies to \( \sin(x) \csc(x) = 1 \).
   - The second term simplifies to \( \cos(x) \csc(x) \cot(x) = \cot(x) \).

7. **Combine Results:**
   \[
   \frac{d}{dx}(-\cos(x) \csc(x)) = 1 + \cot(x)
   \]

### Final Answer:
\[
\frac{d}{dx}(-\cos(x) \csc(x)) = 1 + \cot(x)
\]

Would you like further details or have questions on this derivation?

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Here are five related questions for further exploration:
1. What is the derivative of \( \csc(x) \) and why?
2. How does the product rule apply in other trigonometric derivatives?
3. Can you find the derivative of \( \cos(x) \csc(x) \) without the negative?
4. What are the steps for simplifying trigonometric expressions after differentiation?
5. How do trigonometric identities simplify derivative calculations?

**Tip:** Using trigonometric identities, like \( \sin(x) \csc(x) = 1 \), can simplify derivatives quickly.

Learn how to find the derivative of -cos(x)csc(x) with a detailed step-by-step solution using the product rule and trigonometric differentiation. This problem is ideal for advanced high school calculus students and covers essential calculus concepts and trigonometric identities.

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