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The problem asks for the derivative of the function:

\[
y = \sin(x) \cdot \cot(x)
\]

We will apply the **product rule** to differentiate this function. The product rule states:

\[
\frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x)
\]

Here, \( u(x) = \sin(x) \) and \( v(x) = \cot(x) \).

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

2. **Differentiate \( v(x) = \cot(x) \):**
   The derivative of \( \cot(x) \) is:
   \[
   v'(x) = -\csc^2(x)
   \]

Now apply the product rule:

\[
y' = \cos(x) \cdot \cot(x) + \sin(x) \cdot (-\csc^2(x))
\]

Simplify each term:

\[
y' = \cos(x) \cdot \cot(x) - \sin(x) \cdot \csc^2(x)
\]

We can leave it in this form, but let's compare it to the given options. The closest simplified option is:

\[
\boxed{b. \ -\csc^2(x)}
\]

So, the correct answer is **b**.

Would you like any more details or have further questions?

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Here are 5 related questions to explore further:
1. How is the derivative of \( \cot(x) \) derived?
2. What is the product rule, and how is it applied to more complex functions?
3. How do you find the derivative of a product involving trigonometric functions?
4. How can you use the chain rule in combination with the product rule?
5. Why is the derivative of \( \csc(x) \) useful in calculus?

**Tip:** Memorizing basic trigonometric derivatives (e.g., \( \sin(x) \), \( \cos(x) \), \( \cot(x) \)) helps speed up solving such problems.

Selecciona la opción que corresponde a la derivada de la función y = sen(x) cot(x)

Learn how to find the derivative of the function y = sin(x) * cot(x) using the product rule. This calculus problem involves key trigonometric derivatives and simplification. Suitable for pre-calculus and calculus students.

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