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!