To find the derivative of the function \( y = x \coth(1 + x^2) \), we will apply the product rule and the chain rule.

Given:
\[
y = x \coth(1 + x^2)
\]

### Step 1: Apply the Product Rule
The product rule states:
\[
\frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v'
\]
where \( u = x \) and \( v = \coth(1 + x^2) \).

### Step 2: Differentiate Each Component
1. Differentiate \( u = x \):
   \[
   u' = \frac{d}{dx}(x) = 1
   \]

2. Differentiate \( v = \coth(1 + x^2) \):
   \[
   v' = \frac{d}{dx}(\coth(1 + x^2))
   \]
   To differentiate \( \coth(1 + x^2) \), we use the chain rule. The derivative of \( \coth(u) \) is \( -\csch^2(u) \), so:
   \[
   v' = -\csch^2(1 + x^2) \cdot \frac{d}{dx}(1 + x^2)
   \]
   The derivative of \( 1 + x^2 \) with respect to \( x \) is:
   \[
   \frac{d}{dx}(1 + x^2) = 2x
   \]
   Therefore:
   \[
   v' = -\csch^2(1 + x^2) \cdot 2x
   \]

### Step 3: Combine Using the Product Rule
Now, substitute back into the product rule:
\[
\frac{dy}{dx} = 1 \cdot \coth(1 + x^2) + x \cdot \left(-2x \csch^2(1 + x^2)\right)
\]
Simplifying, we get:
\[
\frac{dy}{dx} = \coth(1 + x^2) - 2x^2 \csch^2(1 + x^2)
\]

So, the derivative is:
\[
\boxed{\frac{dy}{dx} = \coth(1 + x^2) - 2x^2 \csch^2(1 + x^2)}
\]

Would you like more details or have any questions?

Here are five related questions to consider:
1. What is the derivative of \( \coth(x) \) with respect to \( x \)?
2. How would the derivative change if the function was \( y = x \coth(x^2) \) instead?
3. Can you find the second derivative of the function \( y = x \coth(1 + x^2) \)?
4. What are the properties of the hyperbolic cotangent function \( \coth(x) \)?
5. How would you differentiate \( y = \coth(1 + x^2) \) without the \( x \) factor?

**Tip:** Understanding the chain rule is crucial when dealing with composite functions, especially with hyperbolic functions.

Learn how to find the derivative of y = x coth(1 + x^2) using the product rule and chain rule. Detailed step-by-step solution with explanations. Suitable for undergraduate mathematics students.

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