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The function provided is:

\[
y = \sin^2\left(x^3 + 1\right)
\]

We need to differentiate this function with respect to \(x\). This will involve the **chain rule** and **product rule for powers of trigonometric functions**.

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### Steps for differentiation:
1. Rewrite the function:
   \[
   y = \left(\sin\left(x^3 + 1\right)\right)^2
   \]

2. Use the **chain rule**:  
   If \(y = \left(u\right)^2\), where \(u = \sin(x^3 + 1)\), the derivative is:
   \[
   \frac{dy}{dx} = 2u \cdot \frac{du}{dx}
   \]

3. Now, \(u = \sin\left(x^3 + 1\right)\). The derivative of \(u\) with respect to \(x\) is:
   \[
   \frac{du}{dx} = \cos\left(x^3 + 1\right) \cdot \frac{d}{dx}\left(x^3 + 1\right)
   \]

4. Now differentiate \(x^3 + 1\):
   \[
   \frac{d}{dx}(x^3 + 1) = 3x^2
   \]

5. Putting it all together:
   \[
   \frac{du}{dx} = \cos\left(x^3 + 1\right) \cdot 3x^2
   \]

6. Now substitute back into the chain rule formula:
   \[
   \frac{dy}{dx} = 2 \cdot \sin\left(x^3 + 1\right) \cdot \left(\cos\left(x^3 + 1\right) \cdot 3x^2\right)
   \]

7. Simplify:
   \[
   \frac{dy}{dx} = 6x^2 \cdot \sin\left(x^3 + 1\right) \cdot \cos\left(x^3 + 1\right)
   \]

8. Use the **double-angle identity**:  
   \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\), so:
   \[
   \frac{dy}{dx} = 3x^2 \cdot \sin\left(2(x^3 + 1)\right)
   \]

---

### Final Answer:
\[
\frac{dy}{dx} = 3x^2 \cdot \sin\left(2(x^3 + 1)\right)
\]

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Would you like further clarifications or more steps on any part? 

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**Related Questions:**
1. What is the derivative of \(\cos^2(x^2 + 3)\)?
2. How do we apply the product rule for trigonometric functions?
3. Can you explain the chain rule in more detail?
4. What is the significance of double-angle identities in differentiation?
5. How do derivatives change the behavior of trigonometric graphs?

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**Tip:** Memorize basic trigonometric identities—they frequently simplify calculus problems.

Find the derivative of y = sin^2(x^3 + 1)

Find the derivative of the function y = sin^2(x^3 + 1) step-by-step using the chain rule and product rule. This calculus problem covers differentiation techniques for trigonometric functions and includes the application of trigonometric identities.

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