Дана функция Найти f'(0)
F(x) = cosx - x² 1-x 3 f(x)' = (cosx)' - 1/3 (x²)' = (cosx)' (1-x) - cosx (1-x)' - (1/3) * 2x = ( 1-x ) (1-x)² = -(1-x) sinx + cosx - 2x (1-x)² 3 f' (0)= -(1-0) sin 0 + cos 0 - 2*0 = 1/1 =1 (1-0)² 3