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For problem 37, the function given is \( h(x) = 5f(x) - \frac{2}{3}g(x) \), and we are asked to find \( h'(4) \) using the provided table. The derivatives of \( h(x) \) and \( g(x) \) will be needed, as well as their values at \( x = 4 \) from the table.

### Step 1: Differentiate \( h(x) \)
Using the linearity of derivatives, we apply the rules:
\[
h'(x) = \frac{d}{dx}\left(5f(x)\right) - \frac{d}{dx}\left(\frac{2}{3}g(x)\right)
\]
\[
h'(x) = 5f'(x) - \frac{2}{3}g'(x)
\]

### Step 2: Evaluate at \( x = 4 \)
From the table provided:
- \( f'(4) = 3 \)
- \( g'(4) = 4 \)

Substitute these values into the derivative expression:
\[
h'(4) = 5(3) - \frac{2}{3}(4)
\]
\[
h'(4) = 15 - \frac{8}{3}
\]
Now, simplify the expression:
\[
h'(4) = \frac{45}{3} - \frac{8}{3} = \frac{37}{3}
\]

Thus, the value of \( h'(4) \) is:
\[
h'(4) = \frac{37}{3}
\]

Would you like any further clarification or details on these steps?

Here are five follow-up questions that could expand on this problem:
1. How does the chain rule apply when dealing with composite functions in differentiation?
2. What is the product rule, and how would it be used if we had \( f(x)g(x) \)?
3. How do we interpret second derivatives in terms of concavity?
4. What are some common mistakes to avoid when applying the derivative rules?
5. Can the linearity of differentiation be applied to non-linear functions?

**Tip:** When working with derivatives of sums and products, carefully identify which rules apply to each term before proceeding with differentiation.

Use the chart to find h'(4). Given h(x) = 5f(x) - (2/3)g(x).

Solve for h'(4) where h(x) = 5f(x) - (2/3)g(x) using the linearity of derivatives. The solution involves differentiating the function and evaluating it at x = 4 using the given table values. Learn step-by-step how to apply derivative rules.

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