Question

Suppose f(π/3) = 2 and f '(π/3) = −5, and let g(x) = f(x) sin x and h(x) = (cos x)/f(x). Find the following:

g'(π/3)

h'(π/3)

Answer 1

g(x) = f(x) sin x
whats g'(x) from this ?

Answer 2

f'(x)sin(x)+f(x)sin'(x)
I get it (5radical3)/2 +1 it keeps saying that it is wrong

Answer 3

f'(x)sin(x)+f(x)sin'(x)
whats (sin x)' = .... ?

Answer 4

cos(x)

Answer 5

f'(x)sin(x)+f(x)cos x
= -5 sin(pi/3) +2 cos(pi/3)
= -5 sqrt 3/ 2 +2 (1/2) =-5/2 sqrt 3 +1
you are actually correct......hmmm

Answer 6

oh wait!

Answer 7

you missed the -*minus* sign in the beginning :P