can anyone help me with this?

Question

Use the equations and relations given below to answer the following questions.
f(pi/6) = 4 f ' (pi/6) = -6
g(x) = f(x) sin(x) h(x) = (cos(x))/(f(x))

(a) g'(pi/6) =

(b) h'(pi/6) =

Answer 1

For ease of reference, let us number some of the given info. Let

(1) f(pi/6) = -6

(2) f'(pi/6) = -3/2

(3) g(x) = f(x) sin(x) h(x)

(4) g(x) = [cos(x)]/f(x)

(5) f(x) sin(x) h(x) = [cos(x)]/f(x).

From (5) we find g'(x) = {f(x)*[-sin(x)] - [cos(x)]*f'(x)}/[f(x)]^2. Using the known function values, together with sin(pi/6) = 1/2 and cos(pi/6) = sqrt(3)/2, we find g'(pi/6) = [12 + 3sqrt(3)]/4.

By differentiating (3) we find

(6) g'(x) = f'(x) sin(x) h(x) + f(x) cos(x) h(x) + f(x) sin(x) h'(x).

From (5) we get h(x) = [cos(x)]/{[sin(x)]*[f(x)]^2}, so h(pi/6) = [sqrt(3)]/36. Using the known values into (6) we find h'(pi/6).