The function given is:

$y = \sin^2(x^2 + 4)$

We want to find the derivative $y'$. To do this, we will use the chain rule.

Steps:

1. Recognize that $y = \sin^2(u)$, where $u = x^2 + 4$. We need to differentiate this expression.
2. Apply the chain rule: $\frac{dy}{dx} = 2\sin(u) \cdot \frac{d}{dx}[\sin(u)] \cdot \frac{du}{dx}$
3. Now, $u = x^2 + 4$, so $\frac{du}{dx} = 2x$.
4. The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$, so: $\frac{dy}{dx} = 2\sin(x^2 + 4) \cdot \cos(x^2 + 4) \cdot 2x$
5. Simplifying this: $y' = 4x\sin(x^2 + 4)\cos(x^2 + 4)$

This can also be written using the double-angle identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$: $y' = 2x\sin(2(x^2 + 4))$

Final derivative:

$y' = 2x \sin(2(x^2 + 4))$

Would you like further explanation or have any questions?

Here are 5 questions related to this topic:

1. What is the chain rule and when do we apply it?
2. Can you explain how the double-angle identity works in trigonometric functions?
3. How does the derivative of $\sin(u)$ behave when $u$ is a function of $x$?
4. Why did we use the chain rule in this problem, and what would happen if we didn't?
5. What is the geometric meaning of the derivative of trigonometric functions?

Tip: When dealing with trigonometric functions raised to a power, always consider using both the chain rule and trigonometric identities for simplification!