#### Answer

(a) $p'(1) = 1 $
(b) $q'(4) = -1/3$

#### Work Step by Step

(a) We have that $p(x) = f(x)*g(x)$ so
$p(1) = f(1)*g(1) $
$p'(1) = f(1)*g'(1) + g(1)*f'(1) $
watching the graph:
$f(1)= 6$
$g'(1)= -(1.5/3) = -1/2 $, this is negative because the slope is negative.
$g(1)= 4$
$f'(1)= 2/2 = 1$
So
$p'(1) = 6(-1/2) + 4(1) = -3 + 4 = 1$
(b) We have that $q(x) = f(x)/g(x)$ so
$q(4) = f(4)/g(4) $
$q'(4) = \frac{g(4)*f'(4) - f(4)*g'(4)}{[g(4)]^{2}}$
watching the graph:
$g(4)= 3$
$f'(4)= -1$, this is negative because the slope is negative.
$f(4)= 7$
$g'(4)= 0 $
So
$q'(4) = \frac{3(-1) - 7(0)}{9}= \frac{-3}{9} = -1/3$