Calculus 10th Edition

(a) $p'(1) = 1$ (b) $q'(4) = -1/3$
(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 $q'(4) = \frac{3(-1) - 7(0)}{9}= \frac{-3}{9} = -1/3$