Answer
a) $h'(1)$ = $\frac{1}{2}$
b) $s'(5)$ does not exist becuase $g(x)$ is not differentiable at x = 6.
Work Step by Step
a)
$h'(x)$ = $f'(g(x)) $$\times$$ g'(x)$
Looking at the graph, we can see that $g(1)$ = 4. Furthermore, we find that the slope of $g(x)$ at x=1 is $\frac{-1}{2}$, using a secant line which stretches from the points on the graph of g(x) corresponding to the x values of x = -2 and x = 3. Plugging all of these into our equation, we get that $h'(1)$ = $f'(4)$ $\times$ $\frac{-1}{2}$. Using another secant line (which stretches from the values corresponding to the x values of x = 3 and x = 7), we get that the slope of the line at x = 4 of $f(x)$ = -1. Thus, $h'(1)$ = -1 $\times$ $\frac{-1}{2}$ = $\frac{1}{2}$
b)
$s'(x)$ = $g'(f(x))$ $\times f'(x)$
$s'(5)$ = $g'(f(5))$ $\times f'(5)$
Looking at the graphs, we can see that
$f(5)$ = 6, $f'(5)$ = -1
Plugging these values back in, we get
$s'(5)$ = $g'(6)$ $\times$ -1
However, $g'(6)$ does not exist since $g(x)$ is not differentiable at x = 6 (derivatives do not occur at sharp turns). Thus, $s'(5)$ does not exist becuase $g(x)$ is not differentiable at x = 6.
