Let's solve the problem step by step.

We are given:

• A table of values for $f(x)$, $f'(x)$, $g(x)$, and $g'(x)$.
• Functions $f$ and $g$ are continuously differentiable.

We are tasked with solving the following parts:

Part A: Find $h'(2)$ if $h(x) = g(f(x))$.

To find $h'(2)$, use the chain rule: $h'(x) = g'(f(x)) \cdot f'(x).$

Substitute $x = 2$: $h'(2) = g'(f(2)) \cdot f'(2).$

From the table:

• $f(2) = 5$,
• $f'(2) = 1$,
• $g'(5) = 5$ (from the row where $x = 5$).

Substitute these values: $h'(2) = g'(5) \cdot f'(2) = 5 \cdot 1 = 5.$

Answer for Part A: $h'(2) = 5$.

Part B: Find $m'(2)$ if $m(x) = f(x^2)$.

To find $m'(x)$, use the chain rule: $m'(x) = f'(x^2) \cdot \frac{d}{dx}(x^2) = f'(x^2) \cdot 2x.$

Substitute $x = 2$: $m'(2) = f'(2^2) \cdot 2(2) = f'(4) \cdot 4.$

From the table:

• $f'(4) = -6$.

Substitute this value: $m'(2) = (-6) \cdot 4 = -24.$

Answer for Part B: $m'(2) = -24$.

Part C: Let $k(x) = f(g(x))$. Write an equation for the line tangent to the graph of $k$ at $x = 4$.

1. Find $k'(x)$: Use the chain rule: $k'(x) = f'(g(x)) \cdot g'(x).$

2. Find $k'(4)$: Substitute $x = 4$: $k'(4) = f'(g(4)) \cdot g'(4).$

   From the table:

   • $g(4) = 2$,
   • $g'(4) = 5$,
   • $f'(2) = 1$ (since $g(4) = 2$).

   Substitute these values: $k'(4) = f'(2) \cdot g'(4) = 1 \cdot 5 = 5.$

3. Find the point $(4, k(4))$: Substitute $x = 4$ into $k(x)$: $k(4) = f(g(4)).$

   From the table:

   • $g(4) = 2$,
   • $f(2) = 5$.

   Therefore: $k(4) = f(2) = 5.$

4. Equation of the tangent line: The equation of a tangent line is: $y - y_1 = m(x - x_1),$ where $m = k'(4)$, $x_1 = 4$, and $y_1 = k(4)$.

   Substitute: $y - 5 = 5(x - 4).$

   Simplify: $y = 5x - 15.$

Answer for Part C: The equation of the tangent line is $y = 5x - 15$.

Part D: Let $j(x) = \frac{g(x)}{f(x)}$. Find $j'(2)$.

To find $j'(x)$, use the quotient rule: $j'(x) = \frac{f(x) \cdot g'(x) - g(x) \cdot f'(x)}{[f(x)]^2}.$

Substitute $x = 2$: $j'(2) = \frac{f(2) \cdot g'(2) - g(2) \cdot f'(2)}{[f(2)]^2}.$

From the table:

• $f(2) = 5$,
• $g'(2) = 1$,
• $g(2) = 2$,
• $f'(2) = 1$.

Substitute these values: $j'(2) = \frac{(5)(1) - (2)(1)}{(5)^2}.$

Simplify: $j'(2) = \frac{5 - 2}{25} = \frac{3}{25}.$

Answer for Part D: $j'(2) = \frac{3}{25}$.

Summary of Answers:

• Part A: $h'(2) = 5$.
• Part B: $m'(2) = -24$.
• Part C: Tangent line is $y = 5x - 15$.
• Part D: $j'(2) = \frac{3}{25}$.

Do you want further clarification on any part?
Here are 5 related questions you might find helpful:

1. How is the chain rule applied to composite functions like $g(f(x))$?
2. What are the steps for applying the quotient rule to rational functions?
3. Why does $k'(x) = f'(g(x)) \cdot g'(x)$ require the chain rule?
4. How can you verify values from the table for accuracy when solving derivatives?
5. What is the geometric meaning of the tangent line in Part C?

Tip: When working with derivatives, always check if the problem involves composition, product, or quotient rules to apply the correct formula!