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Let's work through both parts of Problem #7 step by step. Here's the breakdown:

### Given:
- Two functions \( f(x) \) and \( g(x) \) with their graphs displayed.
- We are tasked with finding:
  1. \( u'(x) \) at \( x = -3 \), where \( u(x) = f(x)g(x) \).
  2. \( v'(x) \) at \( x = 4 \), where \( v(x) = f(f(x)) \).

---

### Part (a): \( u(x) = f(x)g(x) \), find \( u'(-3) \)

To differentiate \( u(x) = f(x)g(x) \), we apply the **product rule**:
\[
u'(x) = f'(x)g(x) + f(x)g'(x)
\]
We need to evaluate \( u'(-3) \), so we require the values of \( f(-3), g(-3), f'(-3), \) and \( g'(-3) \) from the graph.

From the graph:
- \( f(-3) = 0 \)
- \( g(-3) = -1 \)
- \( f'(-3) \) is the slope of \( f(x) \) at \( x = -3 \). The slope is 1 (since \( f(x) \) is linear at that point).
- \( g'(-3) \) is the slope of \( g(x) \) at \( x = -3 \). From the graph, the slope is approximately 1.

Substituting these values into the product rule:
\[
u'(-3) = f'(-3)g(-3) + f(-3)g'(-3)
\]
\[
u'(-3) = (1)(-1) + (0)(1) = -1
\]
Thus, \( u'(-3) = -1 \).

---

### Part (b): \( v(x) = f(f(x)) \), find \( v'(4) \)

To differentiate \( v(x) = f(f(x)) \), we apply the **chain rule**:
\[
v'(x) = f'(f(x)) \cdot f'(x)
\]
We need to evaluate \( v'(4) \), so we require the values of \( f(4), f'(4), \) and \( f'(f(4)) \).

From the graph:
- \( f(4) = 2 \)
- \( f'(4) \) is the slope of \( f(x) \) at \( x = 4 \). The slope is approximately 1.
- \( f'(f(4)) = f'(2) \), so we need the slope of \( f(x) \) at \( x = 2 \). From the graph, the slope at \( x = 2 \) is 0 (since the graph is flat at this point).

Substituting these values into the chain rule:
\[
v'(4) = f'(f(4)) \cdot f'(4) = f'(2) \cdot f'(4) = 0 \cdot 1 = 0
\]
Thus, \( v'(4) = 0 \).

---

### Summary of answers:
- (a) \( u'(-3) = -1 \)
- (b) \( v'(4) = 0 \)

Would you like more details on any part of the solution, or do you have any questions?

### Relative Questions:
1. What is the product rule for differentiating functions?
2. How do you interpret slopes from the graph of a function?
3. How does the chain rule apply to composite functions like \( f(f(x)) \)?
4. What does the derivative \( u'(x) \) represent in terms of the original functions \( f(x) \) and \( g(x) \)?
5. How would the solution change if the slopes at the required points were different?

### Tip:
When using the product or chain rule, always check the values of the functions and their slopes (derivatives) carefully from the graph, especially at specific points.

Let f and g be the functions whose graphs are shown below. (a) Let u(x) = f(x)g(x). Find u'(-3). (b) Let v(x) = f(f(x)). Find v'(4).

In this problem, we explore how to use the product and chain rules of differentiation for the functions f(x) and g(x) whose graphs are provided. Learn how to find u'(-3) and v'(4) by interpreting slopes from the graph and applying differentiation rules. Suitable for students in Grades 11-12.

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