To find the derivative of $f(x) = x \sin(x)$, we can use the product rule for differentiation, which states that for two functions $u(x)$ and $v(x)$, the derivative of their product $u(x)v(x)$ is:

$(u \cdot v)' = u'v + uv'$

In this case:

• Let $u(x) = x$ and $v(x) = \sin(x)$.
• Then $u'(x) = 1$ and $v'(x) = \cos(x)$.

Using the product rule:

$f'(x) = u'v + uv' = (1) \sin(x) + x \cos(x)$

So,

$f'(x) = \sin(x) + x \cos(x)$

The correct answer is: $f'(x) = \sin(x) + x \cos(x)$

This corresponds to the option: $f'(x) = x \cos(x) + \sin(x)$.

Would you like further details on this process?

Here are 5 related questions to further explore differentiation:

1. How would you differentiate $f(x) = x^2 \sin(x)$ using the product rule?
2. What is the derivative of $f(x) = x \cos(x)$?
3. How does the product rule compare to the chain rule in differentiation?
4. Can you apply the product rule multiple times if you have more than two functions multiplied together?
5. How would you find the second derivative of $f(x) = x \sin(x)$?

Tip: For functions involving products, remember to always apply the product rule carefully to each term.