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.