Suppose f(pi/3) = -6 and f'(pi/2) = 6, and let g(x) = f(x)sinx and h(x) = (cosx)/f(x). Find g'(pi/2) and h'(pi/2)

Question

Suppose f(pi/3) = -6 and f'(pi/2) = 6, and let g(x) = f(x)sinx and h(x) = (cosx)/f(x). Find g'(pi/2) and h'(pi/2)

Answer 1

g(x) = f(x)sin(x)

g'(x) = f '(x)sin(x) + f(x)cos(x)

g'(π/2) = f'(π/2)sin(π/2) + f(x)(cos(π/2))

= 6 (1) + f(π/2)(0)

= 6

h(x) = cos(x)/f(x)

h'(x) = (-sin(x)f(x) - cos(x)f'(x))/f²(x)

h'(π/2) = (-sin(π/2)f(π/2) - cos(π/2)f'(π/2))/f²(x)

= (-1f(π/2) - 0(6))/f²(π/2)

= -1/f(π/2)

Need to find the value of f(π/2)

f'(π/2) = 6 --> f'(x) = 6sin(x) might work

f(x) = -6cos(x) + C

f(π/3) = -6(1/2) +C = -6 --> C = -3

f(x) = -6cos(x) -3

f(π/2) = -3

Therefore,

h'(π/2) = -1/f(π/2) = 1/3

Answer 2

g'(π2)=f'(π/2) from the product rule

To calculate h'(π/2) you need f(π/2) which is not given.