2.
ta có: f(x) = sin^6x + cos^6x
suy ra f'(x) = 6sinx^5.cosx - 6cosx^5.sinx
ta lại có: g(x) = 3sin^2x.cos^2x
suy ra g'(x) = 3.cosx^2.(2.sinx.cos) - 3sinx^2.(2.cosx.sinx)
= 6.sinx.cosx^3 - 6sinx^3.cosx
suy ra f'(x) + g'(x) = 6sinx^5.cosx - 6cosx^5.sinx + 6.sinx.cosx^3 - 6sinx^3.cosx
= 6(sinx^3.cosx.(sinx^2 - 1) - sinx.cosx^3(cosx^2 - 1))
= 6.(sinx^3.cosx.(-cosx^2) - sinx,cosx^3.(-sinx^2))
= 0