Question

Suppose f(π/3) =4 and f’(π/3)=-2, and let g(x)=f(x) sin x and h(x)= (cos x)/f(x). Find
(a) g’(π/3) (b) h’(π/3)

Answer 1

g(x)=f(x) sin x


g'(x)=f'(x) sin(x)+ f(x) cos(x) ... use the product rule here


g'(π/3)=f'(π/3) sin(π/3)+ f(π/3) cos(π/3)


g'(π/3)=(-2)(sqrt(3)/2)+ (4)(1/2)


g'(π/3)=-sqrt(3)+2


I'll let you do the other one

Answer 2

how did you get the -2 and 4

Answer 3

the -2 comes from f’(π/3)=-2 and the 4 comes from f(π/3) =4, both of which are given

Answer 4

oops! my eyes must of just skip that, lol. Thanks!