Let's analyze and solve each part of this problem based on the given information.

1. Identifying $u(x) = f(x) \cdot g(x)$ and finding $u'(x)$: Since $u(x) = f(x) \cdot g(x)$, we use the product rule to find $u'(x)$: $u'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x)$ For $u'(1)$, we need the values of $f(1)$, $g(1)$, $f'(1)$, and $g'(1)$ from the graph.

2. Identifying $v(x) = \frac{f(x)}{g(x)}$ and finding $v'(x)$: Since $v(x) = \frac{f(x)}{g(x)}$, we use the quotient rule to find $v'(x)$: $v'(x) = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}$ For $v'(5)$, we need the values of $f(5)$, $g(5)$, $f'(5)$, and $g'(5)$ from the graph.

Step-by-Step Solutions

(a) Find $u'(1)$:

From the graph:

• $f(1) = 3$
• $g(1) = 1$
• $f'(1)$: The slope of $f(x)$ at $x = 1$ is the slope of the line segment around $x = 1$, which appears to be $-1$.
• $g'(1)$: The slope of $g(x)$ at $x = 1$ is also calculated from the slope of the segment, which appears to be $2$.

Now, plug these values into the formula for $u'(x)$: $u'(1) = f'(1) \cdot g(1) + f(1) \cdot g'(1) = (-1) \cdot 1 + 3 \cdot 2 = -1 + 6 = 5$ So, $u'(1) = 5$.

(b) Find $v'(5)$:

From the graph:

• $f(5) = 2$
• $g(5) = 3$
• $f'(5)$: The slope of $f(x)$ at $x = 5$ is calculated from the nearby segment, which appears to be $-1$.
• $g'(5)$: The slope of $g(x)$ at $x = 5$ is also calculated from the segment, which appears to be $1$.

Now, plug these values into the formula for $v'(x)$: $v'(5) = \frac{f'(5) \cdot g(5) - f(5) \cdot g'(5)}{[g(5)]^2} = \frac{(-1) \cdot 3 - 2 \cdot 1}{3^2} = \frac{-3 - 2}{9} = \frac{-5}{9}$ So, $v'(5) = -\frac{5}{9}$.

Answers

• (a) $u'(1) = 5$
• (b) $v'(5) = -\frac{5}{9}$

Would you like more details, or have questions on any step?

Additional Related Questions:

1. How would the result change if $f(x)$ or $g(x)$ had a different slope at $x = 1$?
2. What if we needed to find $u'(x)$ for another value, such as $x = 2$?
3. How would the derivative change if $u(x)$ were defined as $f(x) + g(x)$ instead of $f(x) \cdot g(x)$?
4. What are the implications if $g(x)$ had a value of zero at $x = 1$?
5. Could we find $v'(x)$ if $g(x)$ were zero at $x = 5$? What would happen?

Tip:

Always verify the slopes (derivatives) carefully by calculating the rise-over-run for each segment on the graph; it ensures precise differentiation.