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To solve part (b) of the problem, where \( G(x) = g(h(f(x))) \), we are asked to find \( G'(1) \).

We will use the **chain rule** to compute \( G'(x) \). Specifically, the chain rule states that:

\[
G'(x) = g'(h(f(x))) \cdot h'(f(x)) \cdot f'(x)
\]

We are tasked with finding \( G'(1) \). So, let's evaluate this step by step using the given table values:

1. **Find \( f(1) \)**:
   From the table, when \( x = 1 \), \( f(1) = 3 \).

2. **Find \( h(f(1)) = h(3) \)**:
   From the table, when \( x = 3 \), \( h(3) = 2 \).

3. **Find \( g(h(f(1))) = g(2) \)**:
   From the table, when \( x = 2 \), \( g(2) = 1 \).

Now, we apply the chain rule derivative:

\[
G'(x) = g'(h(f(x))) \cdot h'(f(x)) \cdot f'(x)
\]

4. **Find \( f'(1) \)**:
   From the table, when \( x = 1 \), \( f'(1) = 3 \).

5. **Find \( h'(f(1)) = h'(3) \)**:
   From the table, when \( x = 3 \), \( h'(3) = 4 \).

6. **Find \( g'(h(f(1))) = g'(2) \)**:
   From the table, when \( x = 2 \), \( g'(2) = 3 \).

Now substitute these into the chain rule:

\[
G'(1) = g'(2) \cdot h'(3) \cdot f'(1) = 3 \cdot 4 \cdot 3 = 36
\]

So, the derivative \( G'(1) = 36 \).

Would you like further details or explanations? Here are 5 related questions for practice:

1. What is the chain rule, and how is it applied in multi-step functions like this one?
2. How would you compute \( F'(2) \) from part (a) of the same problem?
3. If \( G(x) = g(f(h(x))) \), how would you compute \( G'(2) \)?
4. Can you explain the difference between the product rule and chain rule for derivatives?
5. How can we extend the chain rule to functions of several variables?

**Tip**: Always carefully evaluate the inner functions first when applying the chain rule!

Find the derivative G'(1) using the chain rule for the composite function G(x) = g(h(f(x))). Detailed steps using table values for g(x), h(x), and f(x), perfect for calculus students in grades 11-12 or college.

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