Answer
(a) $$F'(2)=20$$
(b) $$G'(3)=63$$
Work Step by Step
(a) $$F(x)=f(f(x))$$
The derivative: $$F'(x)=\frac{df(f(x))}{dx}$$
Apply the Chain Rule, we have $$F'(x)=\frac{df(f(x))}{df(x)}\frac{df(x)}{dx}$$ $$F'(x)=f'(f(x))f'(x)$$
Therefore, $$F'(2)=f'(f(2))f'(2)$$
From the table, we see that $f(2)=1$ and $f'(2)=5$. So, $$F'(2)=f'(1)\times5$$
Again, from the table, $f'(1)=4$. Therefore, $$F'(2)=4\times5=20$$
(b) $$G(x)=g(g(x))$$ The derivative: $$G'(x)=\frac{dg(g(x))}{dx}$$
Apply the Chain Rule: $$G'(x)=\frac{dg(g(x))}{dg(x)}\frac{dg(x)}{dx}$$ $$G'(x)=g'(g(x))g'(x)$$
So, $$G'(3)=g'(g(3))g'(3)$$
From the table, $g(3)=2$ and $g'(3)=9$ $$G'(3)=g'(2)\times9$$
From the table, $g'(2)=7$ $$G'(3)=7\times9=63$$
