https://file.aiquickdraw.com/mathbot/file/1729784477607cbyqc76d.jpg

Let's solve the two parts of the question based on the graph provided.

### (a) \( u(x) = f(x)g(x) \), Find \( u'(-3) \).

To find \( u'(-3) \), we will apply the **product rule** of differentiation, which states:
\[
u'(x) = f'(x)g(x) + f(x)g'(x)
\]
We need to find values for \( f(x) \), \( g(x) \), \( f'(x) \), and \( g'(x) \) at \( x = -3 \).

From the graph:
- \( f(-3) = 0 \) (read from the red line)
- \( g(-3) = 2 \) (read from the blue line)
- \( f'(x) \) is the slope of \( f(x) \). On the interval near \( x = -3 \), the slope of \( f(x) \) is \( \frac{3-0}{-2 - (-3)} = 3 \). So, \( f'(-3) = 3 \).
- \( g'(x) \) is the slope of \( g(x) \). Near \( x = -3 \), the slope of \( g(x) \) is \( \frac{2-(-4)}{-3 - (-4)} = \frac{6}{1} = 6 \). So, \( g'(-3) = 6 \).

Using the product rule:
\[
u'(-3) = f'(-3)g(-3) + f(-3)g'(-3)
\]
Substitute the values:
\[
u'(-3) = (3)(2) + (0)(6) = 6
\]

### (b) \( v(x) = f(f(x)) \), Find \( v'(4) \).

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

From the graph:
- \( f(4) = 3 \)
- \( f'(4) \) is the slope of \( f(x) \) near \( x = 4 \), which is \( \frac{4 - 2}{4 - 2} = 1 \).
- Now, \( f'(f(4)) = f'(3) \). The slope \( f'(3) = \frac{3 - 2}{4 - 3} = 1 \).

Using the chain rule:
\[
v'(4) = f'(f(4)) \cdot f'(4) = f'(3) \cdot f'(4) = (1) \cdot (1) = 1
\]

### Final Answers:
- \( u'(-3) = 6 \)
- \( v'(4) = 1 \)

Would you like further clarification or details on any of these steps? Here are five related questions you might consider:

1. How is the product rule applied in more complex functions?
2. Can the chain rule be used for functions that are more nested than \( f(f(x)) \)?
3. What are the graphical interpretations of derivatives for non-linear functions?
4. How can second derivatives provide more insight into graph behavior?
5. What are some common mistakes when applying the product and chain rules?

**Tip:** When using the product rule, always identify the function and its derivative clearly to avoid mixing up terms during differentiation.

Let u(x) = f(x)g(x). Find u'(-3). Let v(x) = f(f(x)). Find v'(4).

Solve a calculus problem involving the product and chain rule for derivatives of composite functions. The question requires finding u'(-3) and v'(4) for functions f(x) and g(x). Step-by-step solutions are provided, including graph-based slope calculations and applying calculus rules.

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