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.

Question

g'(𝜋/3)=

h'(𝜋/3)=

 

Answer 1

For Part 1:

g'(x) = f'(x)sinx + f(x)cosx; g'(π/3) = f'(π/3)sin(π/3) + f(π/3)cos(π/3) = - 5·√3/2 + 2·1/2 = -5√3/2 + 1

For Part 2:

h'(x) = (- sinxf(x) - cosxf'(x))/f(x)²; h'(π/3) = (- √3/2·2 + 1/2·5)/4 = (5 - 2√3)/8