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To solve the given problem, we need to use the information provided in the graph and apply the rules for differentiation, specifically the product rule and quotient rule.

The functions \( f(x) \) and \( g(x) \) are shown graphically, and we are asked to calculate the derivatives of two new functions:

- \( u(x) = f(x)g(x) \)
- \( v(x) = \frac{f(x)}{g(x)} \)

We will apply the following rules:

1. **Product rule for \( u(x) = f(x)g(x) \)**:
   \[
   u'(x) = f'(x)g(x) + f(x)g'(x)
   \]

2. **Quotient rule for \( v(x) = \frac{f(x)}{g(x)} \)**:
   \[
   v'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}
   \]

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### Part (a) Find \( u'(1) \):

We will need the values and derivatives of \( f(x) \) and \( g(x) \) at \( x = 1 \).

From the graph:
- \( f(1) = 0 \)
- \( g(1) = 2 \)

Now, let's estimate the slopes (derivatives):
- The slope of \( f(x) \) at \( x = 1 \) can be calculated using the points around \( x = 1 \). The slope \( f'(1) \) is the change in \( y \)-values divided by the change in \( x \)-values. From \( x = 0 \) to \( x = 1 \), the function rises from \( f(0) = -1 \) to \( f(1) = 0 \), giving us:
  \[
  f'(1) = \frac{0 - (-1)}{1 - 0} = 1
  \]
- The slope of \( g(x) \) at \( x = 1 \) can be calculated similarly. From \( x = 0 \) to \( x = 1 \), \( g(x) \) rises from \( g(0) = 3 \) to \( g(1) = 2 \), giving us:
  \[
  g'(1) = \frac{2 - 3}{1 - 0} = -1
  \]

Now, apply the product rule:
\[
u'(1) = f'(1)g(1) + f(1)g'(1) = (1)(2) + (0)(-1) = 2
\]

Thus, \( u'(1) = 2 \).

---

### Part (b) Find \( v'(5) \):

We need the values and derivatives of \( f(x) \) and \( g(x) \) at \( x = 5 \).

From the graph:
- \( f(5) = 3 \)
- \( g(5) = 5 \)

For the slopes:
- From \( x = 4 \) to \( x = 5 \), \( f(x) \) falls from \( f(4) = 4 \) to \( f(5) = 3 \), so:
  \[
  f'(5) = \frac{3 - 4}{5 - 4} = -1
  \]
- The slope of \( g(x) \) at \( x = 5 \) is zero, as \( g(x) \) is constant at this point.

Now, apply the quotient rule:
\[
v'(5) = \frac{f'(5)g(5) - f(5)g'(5)}{(g(5))^2} = \frac{(-1)(5) - (3)(0)}{(5)^2} = \frac{-5}{25} = -\frac{1}{5}
\]

Thus, \( v'(5) = -\frac{1}{5} \).

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Would you like further explanation on any part of this solution? Here are some related questions you might find useful:

1. How do you calculate slopes from graphs more accurately?
2. What are the steps to apply the product rule in calculus?
3. Can the quotient rule be used for other types of rational functions?
4. How do the signs of derivatives affect the behavior of functions?
5. What are some examples where both product and quotient rules are needed?

**Tip:** Always carefully estimate slopes from graphs, as they may not be perfectly clear depending on the scale. Use multiple points if needed!

Let u(x) = f(x)g(x) and v(x) = f(x)/g(x). Find u'(1) and v'(5) based on the graph of the functions f and g provided.

Learn how to compute derivatives using the product and quotient rules from the graph of functions f and g. This calculus problem demonstrates finding u'(1) and v'(5) by interpreting slopes and applying key differentiation techniques.

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