The problem is asking to find derivatives of two functions:

1. $p'(9)$ where $p(x) = f(x)g(x)$
2. $q'(4)$ where $q(x) = \frac{f(x)}{g(x)}$

The graph provided shows both $f(x)$ and $g(x)$, so let's use the product and quotient rules for differentiation.

(a) $p'(9)$ – Product Rule:

For $p(x) = f(x)g(x)$, the derivative is given by the product rule: $p'(x) = f'(x)g(x) + f(x)g'(x)$ To calculate $p'(9)$, we need:

• $f(9)$, $f'(9)$
• $g(9)$, $g'(9)$

From the graph at $x = 9$:

• $f(9) = 1$
• $g(9) = 4$

Now, the slopes at $x = 9$ (to find the derivatives):

• $f'(9)$: The graph of $f(x)$ looks flat at $x = 9$, so $f'(9) = 0$.
• $g'(9)$: From the graph, the slope of $g(x)$ at $x = 9$ appears to be $1$.

Now substitute into the product rule: $p'(9) = f'(9)g(9) + f(9)g'(9) = 0 \times 4 + 1 \times 1 = 1$

So, $p'(9) = 1$.

(b) $q'(4)$ – Quotient Rule:

For $q(x) = \frac{f(x)}{g(x)}$, the derivative is given by the quotient rule: $q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$ To calculate $q'(4)$, we need:

• $f(4)$, $f'(4)$
• $g(4)$, $g'(4)$

From the graph at $x = 4$:

• $f(4) = 6$
• $g(4) = 4$

Now, the slopes at $x = 4$:

• $f'(4) = -1$ (decreasing slope)
• $g'(4) = -\frac{1}{2}$ (slightly decreasing slope)

Now substitute into the quotient rule: $q'(4) = \frac{f'(4)g(4) - f(4)g'(4)}{g(4)^2} = \frac{(-1)(4) - (6)(-\frac{1}{2})}{4^2}$ $q'(4) = \frac{-4 + 3}{16} = \frac{-1}{16}$

So, $q'(4) = -\frac{1}{16}$.

Would you like any further details or explanations?

Here are some related questions:

1. What is the general product rule for differentiation?
2. How is the quotient rule derived from the product rule?
3. How do you estimate the slope from a graph visually?
4. What is the importance of checking points and slopes in real-world contexts?
5. How can these differentiation techniques be applied in physics or economics?

Tip: Always ensure your slope estimates from graphs are as precise as possible by looking at intervals between points.