(a) \(P^{\prime}(2)\) 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 \(P^{\prime}(x)\) will look like this:
\(P^{\prime}(x)=F(x) C^{\prime}(x)+F^{\prime}(x) G(x)\)
Now we let \(x=2\).
\(P^{\prime}(2)=F(2) C^{\prime}(2)+F^{\prime}(2) C(2)\)
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 \(C(2)=2\).
\(F^{\prime}(2)\) will be the slope of the curve \(F\) at 2 . A tangent line at \(F(2)\) would be horizontal,
so \(F^{\prime}(2)=0\).
\(G(2)\) has a slope of \(1 / 2\), so \(C^{\prime}(2)=1 / 2\)
Now we substitute these values into \(P^{\prime}(2)\) to solve:
\(P^{\prime}(2)=(3)(1 / 2)+(0)(2)=(3 / 2)+(0)=3 / 2\)
(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.
DIfferentlate \((Q u o t l e n t ~ R u l e):\)
\(Q^{\prime}(x)=\left[G(x) F^{\prime}(x) \cdot F(x) G^{\prime}(x)\right] /\left[G(x)^{2}\right]\)
Read the values off the graph:
\(G(7)=1\)
\(C^{\prime}(7)=-2 / 3\)
\(F(7)=5\)
\(F^{\prime}(7)=1 / 4\)
Substitute:
\(Q^{\prime}(7)=\left[1^{*}(1 / 4)-5^{*}(-2 / 3)\right] /\left[1^{2}\right]\)
\(Q^{\prime}(7)=[(1 / 4)+(10 / 3)]\)
\(Q^{\prime}(7)=(43 / 12)=3.5833\).
