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)]/[1²]
Q’(7)=[(1/4)+(10/3)]
Q’(7)=(43/12)=3.5833.