1) Найдите производную функции в точке Хо: а) y = 2x³, при Хо = -1; 6) y = sinx, Xo = π/3; B)y=-2cosx, Хо=π/4; г) у = 1 +2√x, Xo=9

а) (y = 2x^3). (y' = 6x^2). При (x_0 = -1), (y'(-1) = 6(-1)^2 = 6). б) (y = \sin x). (y' = \cos x). При (x_0 = \frac{\pi}{3}), (y'(\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}). в) (y = -2\cos x). (y' = 2\sin x). При (x_0 = \frac{\pi}{4}), (y'(\frac{\pi}{4}) = 2\sin(\frac{\pi}{4}) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}). г) (y = 1 + 2\sqrt{x}). (y' = \frac{1}{\sqrt{x}}). При (x_0 = 9), (y'(9) = \frac{1}{\sqrt{9}} = \frac{1}{3}).