Calculus
posted by Christina .
Suppose f(pi/3) = 3 and f '(pi/3) = −7,
and let g(x) = f(x) sin x and
h(x) = (cos x)/f(x). Find the following.
a. g'(pi/3)
b. h'(pi/3)

g = f*sin(x)
g' = f'*sin(x) + f*cos(x)
g'(pi/3) = (7)*√3/2 + 3*1/2 = (37√3)/2
h = cos(x)/f
h' = (sin(x)*f  cos(x)*f')/f^2
h'(pi/3) = [(√3/2)*3  (1/2)(7)]/9 = (73√3)/18