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Let's find the derivative of the given function \( y = (5x^2 + 9)\sin(3x) \) using the product rule.

The product rule states:
\[
\frac{d}{dx}(uv) = u'v + uv'
\]
where \( u = 5x^2 + 9 \) and \( v = \sin(3x) \).

### Step-by-step solution:
1. Differentiate \( u = 5x^2 + 9 \):
\[
u' = \frac{d}{dx}(5x^2 + 9) = 10x
\]
2. Differentiate \( v = \sin(3x) \):
\[
v' = \frac{d}{dx}(\sin(3x)) = 3\cos(3x)
\]
3. Now apply the product rule:
\[
\frac{dy}{dx} = u'v + uv'
\]
Substituting the expressions for \( u \), \( u' \), \( v \), and \( v' \):
\[
\frac{dy}{dx} = (10x)\sin(3x) + (5x^2 + 9)(3\cos(3x))
\]
\[
\frac{dy}{dx} = 10x\sin(3x) + 3(5x^2 + 9)\cos(3x)
\]

This matches the third option:
\[
(10x)\sin(3x) + 3(5x^2 + 9)\cos(3x)
\]

Thus, the correct answer is the last option.

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Would you like more details or have any questions? Here are five related questions to explore:

1. How does the product rule differ from the chain rule in differentiation?
2. What would be the second derivative of the given function?
3. How do we approach finding the derivative of trigonometric functions involving angles like \( 3x \)?
4. How would you differentiate a function with a product and quotient together?
5. Can we simplify the expression further after applying the product rule?

**Tip**: Always double-check each step of differentiation to ensure the correct application of the rules, especially when using multiple rules like the product and chain rule together.

Find \( \frac{dy}{dx} \) if \( y = (5x^2 + 9)\sin(3x) \).

Learn how to differentiate the function y = (5x^2 + 9)sin(3x) using the product rule. This problem involves applying differentiation techniques for algebraic and trigonometric functions, suitable for students in Grades 11-12.

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