Given the function \( F(x) = (4x + 5)(3x^2 + 2)^3 \), if you want to find its derivative \( F'(x) \), we will apply the **product rule** and the **chain rule**.

### Step 1: Identify the components
- Let \( u(x) = 4x + 5 \)
- Let \( v(x) = (3x^2 + 2)^3 \)

### Step 2: Compute the derivatives of \( u(x) \) and \( v(x) \)
- The derivative of \( u(x) = 4x + 5 \) is \( u'(x) = 4 \).
- To find \( v'(x) \), we first need to differentiate \( (3x^2 + 2)^3 \). Using the chain rule:

\[
v'(x) = 3 \cdot (3x^2 + 2)^2 \cdot \frac{d}{dx}(3x^2 + 2)
\]

\[
\frac{d}{dx}(3x^2 + 2) = 6x
\]

So,

\[
v'(x) = 3 \cdot (3x^2 + 2)^2 \cdot 6x = 18x(3x^2 + 2)^2
\]

### Step 3: Apply the product rule
The product rule states:

\[
F'(x) = u'(x)v(x) + u(x)v'(x)
\]

Substitute the derivatives:

\[
F'(x) = 4(3x^2 + 2)^3 + (4x + 5) \cdot 18x(3x^2 + 2)^2
\]

### Final Expression for \( F'(x) \):

\[
F'(x) = 4(3x^2 + 2)^3 + 18x(4x + 5)(3x^2 + 2)^2
\]

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

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Would you like further details or explanations on any specific part of this solution?

Here are some additional questions to consider:
1. How would the derivative change if the exponent of \( (3x^2 + 2) \) were different?
2. What happens to the derivative as \( x \) approaches infinity?
3. Can you factor out any common terms in the expression for \( F'(x) \)?
4. How would the product rule apply if there were three terms instead of two?
5. What is the physical interpretation of the derivative in this context?

**Tip:** When dealing with products of functions, always remember to apply the product rule carefully, considering each term's derivative separately.

Learn how to differentiate the function F(x) = (4x + 5)(3x² + 2)³ by applying the product and chain rules. This step-by-step solution is ideal for students in Grades 11-12, covering essential calculus concepts and differentiation techniques.

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