Answer
(a) $$h'(1)=30$$
(b) $$H'(1)=36$$
Work Step by Step
(a) $$h(x)=f(g(x))$$
The derivative of $h$ would be $$h'(x)=\frac{df(g(x))}{dx}$$
Apply the Chain Rule, we have $$h'(x)=\frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}$$ $$h'(x)=f'(g(x))g'(x)$$
Therefore, $$h'(1)=f'(g(1))g'(1)$$
From the table, we see that $g(1)=2$ and $g'(1)=6$ $$h'(1)=f'(2)\times6$$
Again, from the table, $f'(2)=5$ $$h'(1)=5\times6=30$$
(b) $$H(x)=g(f(x))$$ $$H'(x)=\frac{dg(f(x))}{dx}$$
Apply the Chain Rule: $$H'(x)=\frac{dg(f(x))}{df(x)}\frac{df(x)}{dx}$$ $$H'(x)=g'(f(x))f'(x)$$
So, $$H'(1)=g'(f(1))f'(1)$$
From the table, $f(1)=3$ and $f'(1)=4$ $$H'(1)=g'(3)\times4$$
From the table, $g'(3)=9$ $$H'(1)=9\times4=36$$
