Given y, equals, 2, cotangent, squared, left parenthesis, x, right parenthesis, comma*y*=2cot2(*x*), find start fraction, d, y, divided by, d, x, end fraction, .*dxdy*​.

We are given the function:

\[
y = 2 \cot^2(x)
\]

We are tasked with finding the derivative of \( y \) with respect to \( x \), i.e., \(\frac{dy}{dx}\).

### Step-by-Step Solution:

1. **Recall the derivative of cotangent function:**

   The derivative of \( \cot(x) \) with respect to \( x \) is:

   \[
   \frac{d}{dx}[\cot(x)] = -\csc^2(x)
   \]

2. **Use the chain rule:**

   Since we have \( \cot^2(x) \), we use the chain rule. The derivative of \( \cot^2(x) \) is:

   \[
   \frac{d}{dx}[\cot^2(x)] = 2 \cot(x) \cdot \frac{d}{dx}[\cot(x)] = 2 \cot(x) \cdot (-\csc^2(x))
   \]

   This simplifies to:

   \[
   \frac{d}{dx}[\cot^2(x)] = -2 \cot(x) \csc^2(x)
   \]

3. **Differentiate the given equation:**

   Now, differentiate \( y = 2 \cot^2(x) \) with respect to \( x \):

   \[
   \frac{dy}{dx} = 2 \cdot \frac{d}{dx}[\cot^2(x)] = 2 \cdot (-2 \cot(x) \csc^2(x))
   \]

   Simplifying:

   \[
   \frac{dy}{dx} = -4 \cot(x) \csc^2(x)
   \]

Thus, the derivative of \( y \) with respect to \( x \) is:

\[
\frac{dy}{dx} = -4 \cot(x) \csc^2(x)
\]

Would you like more details or have any questions?

Here are 5 follow-up questions:
1. What are some basic identities for cotangent and cosecant functions?
2. How does the chain rule apply when differentiating trigonometric functions?
3. Can you find the second derivative of this function?
4. How does the derivative of \( \cot(x) \) compare with the derivative of \( \tan(x) \)?
5. What are the practical applications of differentiating trigonometric functions?

**Tip:** When differentiating trigonometric functions raised to powers, always apply the chain rule.

Learn how to differentiate y = 2 cot²(x) step-by-step. This solution covers the chain rule, basic trigonometric derivatives, and provides detailed explanations for each step. Suitable for students in advanced high school calculus or undergraduate calculus courses.

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