**Answer:**

**(a)** P’(2) will be the derivative of P(x) evaluated at x=2. So first, we take a derivative.

In this section, you should have learned product rule, so P’(x) will look like this:

P’(x)=F(x)G’(x)+F’(x)G(x)

Now we let x=2.

P’(2)=F(2)G’(2)+F’(2)G(2)

Read these values off the graph.

We see that F(2)=3. (When its x-value is 2, the curve F has a y-value of 3)

We see that G(2)=2.

F’(2) will be the slope of the curve F at 2. A tangent line at F(2) would be horizontal, so F’(2)=0.

G(2) has a slope of 1/2, so G’(2)=1/2

Now we substitute these values into P’(2) to solve:

P’(2)=(3)(1/2)+(0)(2)=(3/2)+(0)=3/2

**(b)** We’re playing by the same rules as above–differentiate the equation, read the values off the graph, substitute the values into the function, and solve.

**Differentiate (Quotient Rule):**

Q’(x)=[G(x)F’(x)-F(x)G’(x)]/[G(x)^{2}]

**Read the values off the graph:**

G(7)=1

G’(7)=-2/3

F(7)=5

F’(7)=1/4

**Substitute:**

Q’(7)=[1*(1/4)-5*(-2/3)]/[1^{2}]

Q’(7)=[(1/4)+(10/3)]

Q’(7)=(43/12)=3.5833.