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The problem you're working on involves the **Chain Rule** in calculus.

Given:

- \( f(x) = x \cdot g(h(x)) \)
- \( g(4) = 2 \)
- \( g'(4) = 3 \)
- \( h(3) = 4 \)
- \( h'(3) = -2 \)

You are asked to find \( f'(3) \).

### Solution:

1. **Differentiate \( f(x) \) using the product rule**:
   \[
   f(x) = x \cdot g(h(x))
   \]
   Using the product rule:
   \[
   f'(x) = 1 \cdot g(h(x)) + x \cdot \frac{d}{dx}[g(h(x))]
   \]
   Now apply the **Chain Rule** to differentiate \( g(h(x)) \):
   \[
   \frac{d}{dx}[g(h(x))] = g'(h(x)) \cdot h'(x)
   \]
   So,
   \[
   f'(x) = g(h(x)) + x \cdot g'(h(x)) \cdot h'(x)
   \]

2. **Substitute \( x = 3 \) into the equation**:
   \[
   f'(3) = g(h(3)) + 3 \cdot g'(h(3)) \cdot h'(3)
   \]

3. **Use the given information**:
   - \( h(3) = 4 \), so \( g(h(3)) = g(4) = 2 \)
   - \( g'(4) = 3 \)
   - \( h'(3) = -2 \)

   Substitute these values:
   \[
   f'(3) = 2 + 3 \cdot 3 \cdot (-2)
   \]
   Simplifying:
   \[
   f'(3) = 2 + 3 \cdot 3 \cdot (-2) = 2 + 9 \cdot (-2) = 2 - 18 = -16
   \]

So, **\( f'(3) = -16 \)**.

### The correct answer is (A) -16.

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Would you like further clarification or more details?

Here are 5 related questions for practice:
1. What is the product rule for derivatives, and how is it applied?
2. How does the chain rule help in differentiating composite functions?
3. If \( f(x) = x^2 g(h(x)) \), how would you find \( f'(x) \)?
4. How do you interpret \( g'(h(x)) \) in the context of the chain rule?
5. Can you provide an example where the chain rule and product rule are used together?

**Tip**: When using the chain rule, always remember to multiply by the derivative of the inside function!

Let f(x) = x * g(h(x)) where g(4) = 2, g'(4) = 3, h(3) = 4, and h'(3) = -2. Find f'(3).

Learn how to apply the Chain Rule and Product Rule to find the derivative of f(x) = x * g(h(x)), where g(4) = 2, g'(4) = 3, h(3) = 4, and h'(3) = -2. This problem is suitable for students in Grades 11-12 or introductory college calculus.

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