To differentiate \( y = x \arctan(5x) \), we will use the product rule and the chain rule.

### Step 1: Differentiate the product
Since \( y = x \arctan(5x) \) is a product of two functions, we apply the product rule:

\[
\frac{d}{dx}[f(x) g(x)] = f'(x) g(x) + f(x) g'(x)
\]

Let:
- \( f(x) = x \)
- \( g(x) = \arctan(5x) \)

Now, we differentiate each part:

- \( f'(x) = \frac{d}{dx}(x) = 1 \)
- For \( g(x) = \arctan(5x) \), we use the chain rule:
  \[
  g'(x) = \frac{d}{dx}[\arctan(5x)] = \frac{1}{1 + (5x)^2} \cdot 5 = \frac{5}{1 + 25x^2}
  \]

### Step 2: Apply the product rule
Now, substitute back into the product rule:

\[
\frac{dy}{dx} = 1 \cdot \arctan(5x) + x \cdot \frac{5}{1 + 25x^2}
\]

Thus, the derivative is:

\[
\frac{dy}{dx} = \arctan(5x) + \frac{5x}{1 + 25x^2}
\]

This is the derivative of \( y = x \arctan(5x) \).

Would you like further details on any step, or have any specific questions?

### Related questions:
1. How do you differentiate other inverse trigonometric functions, like \( \arcsin(x) \)?
2. Can you show how to differentiate \( y = x \ln(5x) \) using the product rule?
3. How does the chain rule work for more complex functions?
4. What happens when differentiating a quotient of two functions instead of a product?
5. How do you apply the product rule when three functions are multiplied together?

### Tip:
When using the product rule, always double-check your derivatives of each individual function before combining them.

Explore how to differentiate y = x arctan(5x) using the product rule and chain rule. This detailed solution covers the application of the product rule for differentiating the product of two functions, and the chain rule for inverse trigonometric functions, suitable for advanced high school or early college students.

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