The expression given is $y = x \sin x$, and we are asked to find $y_n$, which typically refers to the $n$-th derivative of $y$ with respect to $x$.

Let's start by finding the first few derivatives:

1. First derivative ($y_1$): Using the product rule $\frac{d}{dx}(uv) = u'v + uv'$, where $u = x$ and $v = \sin x$: $y_1 = \frac{d}{dx}(x \sin x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x$

2. Second derivative ($y_2$): Differentiating $y_1 = \sin x + x \cos x$ again: $y_2 = \cos x + (1 \cdot \cos x + x \cdot (-\sin x)) = \cos x + \cos x - x \sin x = 2 \cos x - x \sin x$

3. Third derivative ($y_3$): Differentiating $y_2 = 2 \cos x - x \sin x$ again: $y_3 = -2 \sin x - (1 \cdot \sin x + x \cdot \cos x) = -2 \sin x - \sin x - x \cos x = -3 \sin x - x \cos x$

From these initial derivatives, you can observe that the derivatives follow a certain recursive pattern involving sines and cosines with powers of $x$.

Would you like further details on how to extend this pattern for the $n$-th derivative $y_n$? Here are some related questions to extend your understanding:

1. How is the product rule applied for differentiating products of two functions?
2. What general patterns arise when differentiating trigonometric functions multiple times?
3. What methods can be used to simplify the computation of higher-order derivatives?
4. Can the Leibniz rule (general product rule) help in finding higher-order derivatives?
5. How does the recursive nature of trigonometric derivatives manifest in this example?

Tip: When dealing with higher-order derivatives of trigonometric functions, it's useful to recognize repeating cycles in sine and cosine derivatives, which often help simplify expressions.