https://file.aiquickdraw.com/mathbot/file/1728852298798w5gswi1t.jpg

Let's analyze and solve each part of this problem using the provided information from the chart, graphs, and equations.

### Given Information:
1. Functions:
   - \( j(x) = x^3 \)
   - \( k(x) = 5x \)
   - \( n(x) = \frac{1}{x} \)
2. Derivatives provided in the chart for values of \( f(x) \), \( g(x) \), \( h(x) \), and \( p(x) \) along with their derivatives \( f'(x) \), \( g'(x) \), \( h'(x) \), and \( p'(x) \) at specific \( x \) values.

### Solution Outline
I’ll walk through each part step-by-step to provide answers and explanations.

#### (a) Find \( \frac{d}{dx} [m(x) p(x)] \Big|_{x=5} \)
Using the product rule:
\[
\frac{d}{dx} [m(x) p(x)] = m(x) p'(x) + m'(x) p(x)
\]
Evaluate at \( x = 5 \):
- From the table, \( m(5) = 2 \), \( p(5) = -2 \), \( m'(5) = 3 \), \( p'(5) = -1 \).
- Substitute values:
  \[
  m(5) p'(5) + m'(5) p(5) = (2)(-1) + (3)(-2) = -2 - 6 = -8
  \]
**Answer:** \( -8 \)

#### (b) Find \( \frac{d}{dx} [f(x) g(x) + h(x) p(x)] \Big|_{x=1} \)
Differentiate term by term:
\[
\frac{d}{dx} [f(x) g(x) + h(x) p(x)] = f'(x) g(x) + f(x) g'(x) + h'(x) p(x) + h(x) p'(x)
\]
Evaluate at \( x = 1 \):
- From the table, \( f(1) = 3 \), \( g(1) = 2 \), \( h(1) = -1 \), \( p(1) = -2 \), \( f'(1) = 1 \), \( g'(1) = 4 \), \( h'(1) = 3 \), \( p'(1) = 2 \).
- Substitute values:
  \[
  f'(1) g(1) + f(1) g'(1) + h'(1) p(1) + h(1) p'(1) = (1)(2) + (3)(4) + (3)(-2) + (-1)(2) = 2 + 12 - 6 - 2 = 6
  \]
**Answer:** \( 6 \)

#### (c) Find \( \frac{d}{dx} \left[ \frac{x^2 f(x)}{p(x)} \right] \Big|_{x=1} \)
Using the quotient rule:
\[
\frac{d}{dx} \left[ \frac{x^2 f(x)}{p(x)} \right] = \frac{(2x f(x) + x^2 f'(x)) p(x) - x^2 f(x) p'(x)}{[p(x)]^2}
\]
Evaluate at \( x = 1 \):
- From the table, \( f(1) = 3 \), \( p(1) = -2 \), \( f'(1) = 1 \), \( p'(1) = 2 \).
- Substitute values:
  \[
  \frac{(2 \cdot 1 \cdot 3 + 1^2 \cdot 1)(-2) - 1^2 \cdot 3 \cdot 2}{(-2)^2} = \frac{(6 + 1)(-2) - 3 \cdot 2}{4} = \frac{-14}{4} = -3.5
  \]
**Answer:** \( -3.5 \)

#### (d) Find \( \frac{d}{dx} \left[ \sqrt{x \cdot h(x)} + 4 j(x) \right] \Big|_{x=4} \)
Using the chain rule and the product rule:
\[
\frac{d}{dx} \left[ \sqrt{x \cdot h(x)} + 4 j(x) \right] = \frac{1}{2 \sqrt{x h(x)}} (h(x) + x h'(x)) + 4 j'(x)
\]
Evaluate at \( x = 4 \):
- From the table, \( h(4) = -3 \), \( h'(4) = -2 \), \( j(x) = x^3 \Rightarrow j'(x) = 3x^2 \), so \( j'(4) = 3 \cdot 4^2 = 48 \).
- Substitute values:
  \[
  \frac{1}{2 \sqrt{4 \cdot (-3)}} ( -3 + 4 \cdot -2 ) + 4 \cdot 48
  \]
  **Note:** Here, \( \sqrt{4 \cdot (-3)} \) yields a complex value due to a negative radicand, which typically isn't allowed in a real-value calculus context. Verify this in class if needed.

#### Follow-Up Questions
1. Would you like further details on any specific part?
2. Do you want explanations for each derivative rule applied?
3. Would you like to discuss the graph's role in solving any part?
4. Should I proceed with the remaining questions in detail?
5. Do you need help with interpreting the notation in these problems?

**Tip:** When using product and quotient rules, write down all values for each function and its derivative at a given \( x \)-value to avoid errors in substitution.

Use the values in the chart to evaluate the derivatives of the combinations of functions at given x-values. Show your thinking/work on each one.

Learn how to solve derivative problems using product, quotient, and chain rules. This problem uses values from charts and involves differentiating combinations of functions. Suitable for advanced high school calculus students.

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