Answer
$a) P'(2)=\frac{3}{2}$
$b) Q'(7)=\frac{43}{12}$
Work Step by Step
$$P'(x)=F'(x)G(x)+F(x)G'(x)$$
$$P'(2)=F'(2)G(2)+F(2)G'(2)$$
At $x=2$,$F(2)=3$ and the graph of $F$ has a horizontal asymptote so $F'(2)=0$.
At $x=2$,$G(2)=2$ the value of $G'(2)$ is the slope of the tangent line to $G$ at $x=2$:
$$G'(2)=\frac{2-1}{2-0}=\frac{1}{2}$$
so:
$$P'(2)=0\cdot 2+3\times\frac{1}{2}=\frac{3}{2}$$
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$$Q'(7)=\frac{F'(7)G(7)-F(7)G'(7)}{G^{2}(7)}$$
At $x=7$,$F(7)=5$ and the slope of the tangent line to $F$ at $7$ is:
$$F'(7)=\frac{5-4}{7-3}=\frac{1}{4}$$
At $x=7$,$G(7)=1$ and the slope of the tangent line to $G$ at $7$ is:
$$G'(7)=\frac{1-3}{7-4}=\frac{-2}{3}$$
so:
$$Q'(7)=\frac{\frac{1}{4}\cdot 1-5\cdot (\frac{-2}{3})}{1^{2}}=\frac{43}{12}$$
