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The function provided is \( f(x) = x \sin(x) \). We are asked to find the derivative \( f'(x) \).

To find the derivative, we will use the **product rule**, which states:

\[
(fg)' = f'g + fg'
\]

In this case:
- \( f(x) = x \)
- \( g(x) = \sin(x) \)

Let's apply the product rule:

1. The derivative of \( f(x) = x \) is \( f'(x) = 1 \).
2. The derivative of \( g(x) = \sin(x) \) is \( g'(x) = \cos(x) \).

Now applying the product rule:

\[
f'(x) = (x)' \sin(x) + x (\sin(x))'
\]

\[
f'(x) = 1 \cdot \sin(x) + x \cdot \cos(x)
\]

\[
f'(x) = \sin(x) + x \cos(x)
\]

Thus, the derivative is:

\[
f'(x) = \sin(x) + x \cos(x)
\]

Let me know if you'd like further clarification or details!

Here are 5 related questions to expand your understanding:
1. How would the derivative change if the function was \( f(x) = x^2 \sin(x) \)?
2. Can you apply the product rule to \( f(x) = x \cos(x) \)?
3. How do you find the second derivative of \( f(x) = x \sin(x) \)?
4. How does the chain rule apply to trigonometric functions like \( \sin(x) \)?
5. What is the general form of the product rule for three functions?

**Tip:** The product rule is essential when differentiating the product of two functions. Always identify the two parts and apply the rule carefully!

Learn how to find the derivative of the function f(x) = x sin(x) using the product rule. This solution covers key differentiation concepts and includes step-by-step instructions. Suitable for students in Grades 11-12.

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