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The problem in the image is asking to find the derivative of the function \( h(x) = \sin^2(3x + 1) \).

This is equivalent to \( h(x) = \left( \sin(3x + 1) \right)^2 \).

To solve this, we can use the **chain rule**.

### Steps to find \( h'(x) \):

1. **Outer Function**: The outer function is \( f(u) = u^2 \), where \( u = \sin(3x + 1) \). The derivative of \( u^2 \) with respect to \( u \) is \( 2u \).
   
   So, \( \frac{d}{dx} \left( \sin(3x + 1)^2 \right) = 2 \sin(3x + 1) \cdot \frac{d}{dx} \left( \sin(3x + 1) \right) \).

2. **Inner Function**: Now, take the derivative of the inner function \( \sin(3x + 1) \) with respect to \( x \). Using the chain rule again:
   
   \[
   \frac{d}{dx} \left( \sin(3x + 1) \right) = \cos(3x + 1) \cdot \frac{d}{dx}(3x + 1) = \cos(3x + 1) \cdot 3.
   \]

3. **Final Derivative**: Now, multiplying these results together:
   
   \[
   h'(x) = 2 \sin(3x + 1) \cdot 3 \cos(3x + 1) = 6 \sin(3x + 1) \cos(3x + 1).
   \]

This can be simplified using the double-angle identity for sine: 
\[
\sin(2\theta) = 2 \sin(\theta) \cos(\theta).
\]
So, 
\[
h'(x) = 3 \sin(2(3x + 1)) = 3 \sin(6x + 2).
\]

### Final Answer:
\[
h'(x) = 3 \sin(6x + 2).
\]

Would you like further details or have any other questions?

Here are 5 related questions:
1. How do you apply the chain rule for more complex functions?
2. What is the derivative of \( \cos^2(3x+1) \)?
3. How can you apply trigonometric identities to simplify derivatives?
4. What is the general derivative of \( \sin^n(x) \)?
5. How do you differentiate \( \sin(ax + b)^n \)?

**Tip**: Always check if trigonometric identities can simplify your final derivative!

Tentukan turunan dari h(x) = sin^2(3x + 1) = (sin(3x + 1))^2

Learn how to differentiate the function h(x) = sin^2(3x + 1) using the chain rule and trigonometric identities. This problem involves taking the derivative of a composite function and applying the double angle identity to simplify the final result. Suitable for high school students in Grades 10-12.

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