Найти производную функции в точке xo 1) y=3x'2, xo=1 2) y=cosx, xo=П/6 3) y=-2sin x, co=П/4 4) y=2+корень х, хо=4

1) y' = (3x^2) ' = 6x.

Подставим x0 = 1:

(y (1)) ' = 6 * 1 = 6.

2) y' = (cos (x)) ' = - sin (x).

x0 = π/6.

(y (π/6)) ' = - sin (π/6) = - 1/2.

3) y' = (-2sin (x)) ' = - 2cos (x).

x0 = π/4;

(y (π/4)) ' = - 2cos (π/4) = - 2 * √2/2 = - √2.

4) (y) ' = (2 + √x) ' = 1 / (2√x).

x0 = 4;

(y (4)) ' = 1 / (2√4) = 1/4.