Cho hàm số f(x)=cos^2x+cos^2(2pi/3+x)+cos^2(2pi/3-x). Tính đạo hàm f'(x)

Có f'x=cos2x+cos22π3+x+cos22π3−x'

=cos2x'+cos22π3+x'+cos22π3−x'

=2cosx⋅cosx'+2cos2π3+x⋅cos2π3+x'+2cos2π3−x⋅cos2π3−x'

=−2cosx⋅sinx−2cos2π3+xsin2π3+x+2cos2π3−xsin2π3−x

=−sin2x−sin4π3+2x+sin4π3−2x

=−sin2x−sinπ+π3+2x+sinπ+π3−2x

=−sin2x+sinπ3+2x−sinπ3−2x=−sin2x+2cosπ3sin2x=−sin2x+sin2x=0

Vậy f'(x) = 0 với mọi x Î ℝ.