Пожалуйста помогите найти производные функций 1) f(x)=2x^3-4 x^2+3x-2 2) f(x)=1/3cosx+2tgx-4 3) f(x)=(1/3x^3+6x^2+1)

1) f (x) = 2 * x^3 - 4 * x^2 + 3 * x - 2 ;f \' (x) = (2x^3 - 4x^2 + 3x - 2) \' = (2x^3) \' - (4x^2) \' + (3x) \' - 2 \' = 2 * 3 * x^2 - 4 * 2 * x + 3 = 6x^2 - 8x + 3 . 2) f (x) = 1/3 * cos x + 2 * tg x - 4 ;f \' (x) = (1/3cos x + 2tg x - 4) \' = (1/3cos x) \' + (2tg x) \' - 4 \' = - 1/3sin x + 2 / (cos x)^2 .3) f (x) = (1/3 * x^3 + 6 * x^2 + 1) * (3 * x + 5) ;f \' (x) = ((1/3 * x^3 + 6 * x^2 + 1) * (3 * x + 5)) \' = (1/3 * x^3 + 6 * x^2 + 1) \' * (3 * x + 5) + (1/3 * x^3 + 6 * x^2 + 1) * (3 * x + 5) \' = (1/3 * 3 * x^2 + 6 * 2 * x) * (3 * x + 5) + (1/3 * x^3 + 6 * x^2 + 1) * 3 = (x^2 + 12x) * (3x + 5) + (x^3 + 18x^2 + 3) = 3x^3 + 5x^2 + 36x^2 + 60x + x^3 + 18x^2 + 3 = 4x^3 + 59x^2 + 60x + 3 . 4) f (x) = x^3 + 4/7 * x - 1 ; f \' (x) = (x^3 + 4/7 * x - 1) \' = (x^3) \' + (4/7 * x) \' - 1 \' = 3* x^2 + 4/7 - 0 = 3x^2 + 4/7 .