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To differentiate \( y = x \sin x \), we can apply the **product rule**, which is used when differentiating a product of two functions.

The product rule states that if \( y = u(x) \cdot v(x) \), then:

\[
\frac{dy}{dx} = u'(x) v(x) + u(x) v'(x)
\]

Here, we have:
- \( u(x) = x \)
- \( v(x) = \sin x \)

First, differentiate each function:
- \( u'(x) = 1 \) (the derivative of \( x \))
- \( v'(x) = \cos x \) (the derivative of \( \sin x \))

Now apply the product rule:
\[
\frac{dy}{dx} = u'(x) v(x) + u(x) v'(x) = 1 \cdot \sin x + x \cdot \cos x
\]
\[
\frac{dy}{dx} = \sin x + x \cos x
\]

Looking at the options in the image, the correct answer is:

**Option e.** \( \frac{dy}{dx} = \sin x + x \cos x \).

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Would you like more details on the differentiation process or have any other questions? Here are some related questions to expand on the topic:

1. How does the product rule apply to more complex functions?
2. What is the chain rule, and when do we use it in differentiation?
3. How would we differentiate a function like \( y = x^2 \sin x \)?
4. Can you explain how the derivative of \( \sin x \) becomes \( \cos x \)?
5. What is the general rule for differentiating trigonometric functions?

**Tip:** When differentiating, always check if you're dealing with a product, quotient, or composite functions to apply the correct rule.

Learn how to differentiate y = x sin x by applying the product rule. This calculus problem focuses on using key differentiation techniques, including the product rule and trigonometric derivatives. Suitable for high school students in Grades 11-12.

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