Confusing calculus problem

Number 85:
photobucket.com/albums/
z363/EvangelineChrys/photo.jpg

thanks!

p(x) = f(x)g(x)
p '(x) = f '(x)g(x) + f(x)g '(x)
p '(1) = f '(1)g(1) + f(1)g '(1)
p '(1) = 1(4) + 6(-1/2)
p '(1) = 4 - 3 = 1

q(x) = f(x)/g(x)
q '(x) = (g(x)f '(x) - f(x)g '(x))/(g(x))^2
q '(4) = (3(-1) - 7(0))/3^2 = -3/9 = -1/3