Jika f(x) = cos^2(sin 2x), maka f'(x)= ...

misal u = sin 2x
v = cos u
f(v) = v²
du/dx = 2 sin 2x cos 2x = cos 4x
dv/du = -sinu
df(v)/dv = 2v

df(x)/dv * dv/du * du/dx = 2v*(-sin u)*(cos 4x)
f'(x) = 2 cos(sin2x) * - sin(sin2x) *cos 4x
f'(x) = -sin(2sin2x)*cos4x