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To find the derivative \(\frac{dy}{dx}\) of \( y = (3x^2 + 11) \sin(8x) \), we need to apply the product rule. The product rule states that if we have a function \(y = u(x)v(x)\), then its derivative is given by:

\[
\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)
\]

In this case, let:
\( u(x) = 3x^2 + 11 \) and \( v(x) = \sin(8x) \).

First, we compute the derivatives of \( u(x) \) and \( v(x) \):

1. \( u'(x) = \frac{d}{dx}(3x^2 + 11) = 6x \)
2. \( v'(x) = \frac{d}{dx}(\sin(8x)) \)

To find \( v'(x) \), we use the chain rule:
\[
\frac{d}{dx}(\sin(8x)) = \cos(8x) \cdot \frac{d}{dx}(8x) = 8 \cos(8x)
\]

Now, apply the product rule:
\[
\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)
\]
\[
\frac{dy}{dx} = (6x)\sin(8x) + (3x^2 + 11)(8\cos(8x))
\]

Therefore, the derivative is:
\[
\frac{dy}{dx} = (6x)\sin(8x) + (3x^2 + 11)(8\cos(8x))
\]

This matches the third option in the provided choices. So the correct answer is:
\[
\boxed{(6x)\sin(8x) + (3x^2 + 11)(8\cos(8x))}
\]

Do you want details or have any questions?

Here are 5 related questions to expand your understanding:
1. How do you apply the product rule for differentiation?
2. What is the chain rule in calculus, and how is it used in differentiation?
3. How do you differentiate trigonometric functions such as \(\sin(x)\) and \(\cos(x)\)?
4. Can you explain the steps to find the derivative of a composite function?
5. How would the derivative change if the original function included an additional term?

Tip: When differentiating a product of two functions, always remember to use the product rule, which involves differentiating each function separately and then combining the results.

Find the derivative of y = (3x^2 + 11)sin(8x).

Learn how to differentiate the function y = (3x^2 + 11)sin(8x) using the product rule and chain rule. This step-by-step guide covers key calculus concepts, including trigonometric derivatives. Ideal for students in Grade 11-12 or first-year college calculus.

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