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The provided expression to differentiate is \( G(x) = (3x^2 + 8)(3x + \sqrt{x}) \).

To differentiate \( G(x) \), we can apply the **product rule** of differentiation, which states:

\[
\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)
\]

Let:
- \( f(x) = 3x^2 + 8 \)
- \( g(x) = 3x + \sqrt{x} \)

### Step-by-step differentiation:

1. Differentiate \( f(x) = 3x^2 + 8 \):
   \[
   f'(x) = 6x
   \]

2. Differentiate \( g(x) = 3x + \sqrt{x} \):
   Recall that \( \sqrt{x} = x^{1/2} \), so:
   \[
   g'(x) = 3 + \frac{1}{2}x^{-1/2} = 3 + \frac{1}{2\sqrt{x}}
   \]

Now apply the product rule:

\[
G'(x) = f'(x)g(x) + f(x)g'(x)
\]

Substitute the functions and their derivatives:

\[
G'(x) = (6x)(3x + \sqrt{x}) + (3x^2 + 8)(3 + \frac{1}{2\sqrt{x}})
\]

Next, we simplify each term:

- First term:
\[
(6x)(3x + \sqrt{x}) = 18x^2 + 6x\sqrt{x}
\]

- Second term:
\[
(3x^2 + 8)(3 + \frac{1}{2\sqrt{x}}) = (3x^2 + 8)(3) + (3x^2 + 8)\left(\frac{1}{2\sqrt{x}}\right)
\]
Expanding:
\[
= 9x^2 + 24 + \frac{3x^2}{2\sqrt{x}} + \frac{8}{2\sqrt{x}} = 9x^2 + 24 + \frac{3x^{3/2}}{2} + \frac{4}{\sqrt{x}}
\]

Finally, combine all terms:
\[
G'(x) = 18x^2 + 6x\sqrt{x} + 9x^2 + 24 + \frac{3x^{3/2}}{2} + \frac{4}{\sqrt{x}}
\]

This is the derivative of the function \( G(x) \).

Would you like to see further simplifications or explanations? Also, here are some related questions for you to explore:

1. How does the product rule differ from the quotient rule?
2. Can the expression for \( G'(x) \) be simplified further by combining like terms?
3. What is the significance of the chain rule in cases involving nested functions?
4. How would the derivative change if we had additional terms in \( G(x) \)?
5. What happens when we differentiate a product of more than two functions?

**Tip:** Always break complex expressions into smaller parts when applying the product rule to make the process clearer.

Differentiate G(x) = (3x^2 + 8)(3x + √x) using the product rule in calculus.

Learn how to differentiate G(x) = (3x^2 + 8)(3x + √x) using the product rule in calculus. This detailed solution applies the product rule and other differentiation techniques, including the power rule and simplification. Ideal for students studying calculus at a college level.

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