Найти Vx : y=x^3+x, y=1, x=2

1) y = 6x - 11

y' = 6

2) y = x - 1/2

y' = 1

3) y = x^2 + sinx

y' = 2x + cosx

y' (x0) = 2*pi + cos (pi) = 2*pi - 1

4) y = (x^4) / 2 - (3*x^2) / 2 + 2x

y' = 1/2 * 4x^3 - 1/2 * 6x + 2 = 2x^3 - 3x + 2

y' (x0) = 2*8 - 3*2 + 2 = 16 - 6 + 2 = 12

5) y = sin (3x-2)

y' = cos (3x-2) * (3x-2) ' = 3cos (3x-2)

6) не поняла, что знак "V" обозначает, пусть будет делением

y = 3x^2 - 12/x

y' = 6x - 12 * (-1 / (x^2)) = 6x + 12 / (x^2)

y' (x0) = 6*4 + 12/16 = 24 + 3/4 = 24,75

7) 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).