Let's analyze the graphs of the two functions $f(x)$ (blue) and $g(x)$ (red) to find the values of $u'(1)$ and $v'(1)$.

Given:

• $u(x) = f(x) \cdot g(x)$
• $v(x) = \frac{f(x)}{g(x)}$

We need to use the product rule for $u'(x)$ and the quotient rule for $v'(x)$ at $x = 1$.

1. Find $u'(x)$:

Using the product rule: $u'(x) = f'(x)g(x) + f(x)g'(x)$ We will need $f(1)$, $f'(1)$, $g(1)$, and $g'(1)$.

2. Find $v'(x)$:

Using the quotient rule: $v'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$

Step 1: Derivatives and Function Values at $x = 1$

From the graph:

• $f(1) = 0$ (since the blue graph passes through $(1, 0)$).
• $g(1) = 0$ (since the red graph also passes through $(1, 0)$).

Slopes:

• $f'(1)$ is the slope of the blue line at $x = 1$, which is approximately $1$ (from $(0, 1)$ to $(1, 0)$).
• $g'(1)$ is the slope of the red line at $x = 1$, which appears to be around $2$ (based on the nearby points $(0, -2)$ and $(1, 0)$).

Step 2: Compute $u'(1)$

Substituting into the product rule: $u'(1) = f'(1) \cdot g(1) + f(1) \cdot g'(1)$ $u'(1) = 1 \cdot 0 + 0 \cdot 2 = 0$

Step 3: Compute $v'(1)$

Substituting into the quotient rule: $v'(1) = \frac{f'(1)g(1) - f(1)g'(1)}{g(1)^2}$ $v'(1) = \frac{1 \cdot 0 - 0 \cdot 2}{0^2}$ This expression is undefined because of division by zero. Therefore, $v'(1)$ is undefined.

Final Answer:

• $u'(1) = 0$
• $v'(1)$ is undefined due to division by zero.

Would you like more details or further explanation?

Here are 5 related questions for further exploration:

1. How is the product rule derived for differentiating products of functions?
2. Why does the quotient rule involve the square of the denominator in its formula?
3. What happens graphically when a function has a derivative of 0 at a certain point?
4. What is the significance of slopes (derivatives) for functions at given points?
5. How do we interpret division by zero in a quotient of functions?

Tip: Always check for points where a function or its derivative might be undefined, especially in cases involving division.