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The problem you're working on involves two graphs of functions \( f(x) \) (red) and \( g(x) \) (blue), and asks for derivatives related to the compositions and products of these functions. Here's how we can approach both parts of the problem:

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

We use the product rule for differentiation:
\[
u'(x) = f'(x)g(x) + f(x)g'(x)
\]
We need to find \( u'(3) \), which requires the values of \( f(3) \), \( g(3) \), \( f'(3) \), and \( g'(3) \).

From the graph:
- \( f(3) \) is approximately 4.
- \( g(3) \) is approximately 0 (as the blue line crosses the x-axis near \( x = 3 \)).
- The slope of \( f(x) \) at \( x = 3 \), which is \( f'(3) \), is positive and appears to be around 1 (steepness of red line).
- The slope of \( g(x) \) at \( x = 3 \), which is \( g'(3) \), appears to be slightly negative, around \(-0.5\) (since the blue line is sloping downward).

Thus:
\[
u'(3) = f'(3)g(3) + f(3)g'(3) = (1)(0) + (4)(-0.5) = -2
\]

So, \( u'(3) = -2 \).

---

### (b) Let \( v(x) = f(f(x)) \). Find \( v'(-2) \).

This requires the chain rule:
\[
v'(x) = f'(f(x)) \cdot f'(x)
\]
We need to find \( v'(-2) \), which requires knowing \( f(-2) \), \( f'(-2) \), and \( f'(f(-2)) \).

From the graph:
- \( f(-2) \) is approximately \(-2\).
- The slope of \( f(x) \) at \( x = -2 \), \( f'(-2) \), appears to be about 1.
- Since \( f(-2) = -2 \), we also need \( f'(-2) \) for the inner part of the chain rule. So \( f'(f(-2)) = f'(-2) \), which we already said is about 1.

Thus:
\[
v'(-2) = f'(f(-2)) \cdot f'(-2) = 1 \cdot 1 = 1
\]

So, \( v'(-2) = 1 \).

---

### Do you need any further clarification or details?

Here are some related questions:
1. How do you determine the derivative at a specific point from a graph?
2. Why is the chain rule essential in part (b) for composition of functions?
3. Could the product rule be used for any product of functions, and how is it applied in this case?
4. What would change if the slope of \( g(x) \) at \( x = 3 \) was zero?
5. How would the problem differ if \( v(x) \) was \( g(f(x)) \) instead?

**Tip**: Always check the slopes carefully when working from graphs to determine the derivative values accurately.

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'(-2).

Solve a calculus problem involving derivatives of product and composite functions. Learn how to apply the product rule and chain rule using graphs of f(x) and g(x). This detailed explanation is suitable for students in Grade 12 or first-year college calculus.

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