Jika f(x)= sinxcos3x, maka f'(1/6Π) =...? Tolong beserta caranya,

Turunan

f(x)= sin x cos 3x
f(x) = 1/2 (2 cos 3x sin x)
f(x) = 1/2 [ sin(3x+x) - sin(3x-x)]
f(x) = 1/2 [sin 4x - sin 2x]

f ' (x) = 1/2 [ 4 cos 4x - 2 cos 2x ]
f '(π/6)= 1/2 [ 4 cos 2/3 π - 2 cos 1/3 π]
f '(π/6)= 1/2 [ 4(-1/2) - 2(1/2)]
f '(π/6) = 1/2 [-2 -1] = - 3/2