Suppose f(π/3) = 5 and f '(π/3) = −3, and let g(x) = f(x) sin x

Question

and h(x) = (cos x)/f(x). Find the following. (a) g'(π/3)?

Answer 1

(a)

The function g(x) = f(x) sin x .

differentiate the function with respect to x .

 g'(x) = d/dx( f(x) sin x )

Use product rule of derivatives    uv = uv’ + vu’

g'(x) =  f(x) cos x +  f '(x) sin x           [ derivative of sin x is cos x ]

g'(x) at x = π/3

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

Given  f(π/3) = 5 and  f '(π/3) = -3 .

g'(π/3) = 5 cos π/3 - 3 sin π/3

g'(π/3) = 5(1/2) - 3 √(3) ÷ 2

g'(π/3) = [5 - 3√(3)]/2

So g'(π/3) = [5 - 3√(3)]/2