1. Найдите производную функции в точке x0: а) y=2x², x0=-1 б) y=-2cosx? x0=π/4 в) y=sinx, x0=π/3 г) y=1+2√x, x0=9

а) (y) ' = (2x^2) ' = 4x;

(y (-1)) ' = 4 * (-1) = - 4.

б) (y) ' = (-2cos (x)) ' = - 2 * (-sin (x) = - 2sin (x);

(y (π/4)) ' = - 2 * √2/2 = - √2.

в) (y) ' = (sin (x)) ' = cos (x);

(y (π/3)) ' = cos (π/3) = 1/2.

г) (y) ' = (1+2√x) = 2 * 1/2 * x^ (-1/2) = 1/√x;

(y (9)) ' = 1/√9 = 1/3.