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The function given is:

\[
g(\alpha) = 10 \sin^5(3\alpha)
\]

To find the derivative, \( g'(\alpha) \), we can proceed with the chain rule. Let's break it down step-by-step.

1. **Rewrite the function** for clarity:
   \[
   g(\alpha) = 10 \left(\sin(3\alpha)\right)^5
   \]

2. **Differentiate** using the chain rule. Set \( u = \sin(3\alpha) \), so the function becomes \( g(\alpha) = 10 u^5 \).

3. Differentiate \( 10 u^5 \) with respect to \( u \):
   \[
   \frac{d}{du} \left(10 u^5\right) = 50 u^4
   \]

4. Now differentiate \( u = \sin(3\alpha) \) with respect to \( \alpha \):
   \[
   \frac{d}{d\alpha} (\sin(3\alpha)) = 3 \cos(3\alpha)
   \]

5. Combine these results using the chain rule:
   \[
   g'(\alpha) = 50 \sin^4(3\alpha) \cdot 3 \cos(3\alpha)
   \]

6. Simplify the expression:
   \[
   g'(\alpha) = 150 \sin^4(3\alpha) \cos(3\alpha)
   \]

So, the derivative of \( g(\alpha) = 10 \sin^5(3\alpha) \) is:

\[
g'(\alpha) = 150 \sin^4(3\alpha) \cos(3\alpha)
\]

Would you like a more detailed breakdown of any step?

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**Related Questions:**

1. How do you find the second derivative of this function?
2. What is the derivative if the function were \( g(\alpha) = 10 \cos^5(3\alpha) \)?
3. How would you differentiate \( g(x) = 10 \sin(3x)^5 \) with respect to \( x \)?
4. How does the chain rule apply when differentiating nested trigonometric functions?
5. What is the critical point of this function?

**Tip:** When differentiating trigonometric functions raised to a power, remember to apply the chain rule carefully to each layer.

Find the derivative of g(α) = 10 sin^5(3α).

This solution provides a step-by-step approach to finding the derivative of the function g(α) = 10 sin^5(3α). The solution includes applying the chain rule and differentiating trigonometric functions, making it suitable for high school calculus students in Grades 11-12.

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