Найти Vx : y=x^3+x, y=1, x=2

1) y = 6x - 11y' = 62) y = x - 1/2y' = 13) y = x^2 + sinxy' = 2x + cosxy'(x0) = 2*pi + cos(pi) = 2*pi - 14) y = (x^4)/2 - (3*x^2)/2 + 2xy' = 1/2 * 4x^3 - 1/2 * 6x + 2 = 2x^3 - 3x + 2y'(x0) = 2*8 - 3*2 + 2 = 16 - 6 + 2 = 125) y = sin(3x-2)y' = cos(3x-2)*(3x-2)' = 3cos(3x-2)6) не поняла, что знак "V" обозначает, пусть будет делениемy = 3x^2 - 12/xy' = 6x - 12*(-1/(x^2)) = 6x + 12/(x^2)y'(x0) = 6*4 + 12/16 = 24 + 3/4 = 24,757) y = 1/(2tg(4x-pi)) + pi/4 y' = -1/(2tg^2(4x-pi)) * 1/cos^2(4x-pi) * 4 + 0 = -2/(tg^2(4x-pi)*cos^2(4x-pi)) = -2/sin^2(4x-pi).