Вычислите значение производной функции...

Решите задачу:

$1)\quad y=\frac{1}{2}tg(4x-\pi )+\frac{\pi}{4}\; ,\; x_0=\frac{\pi}{4}\\\\y'=\frac{1}{2}\cdot \frac{1}{cos^2(4x-\pi )}\cdot 4=\frac{2}{cos^2(4x-\pi )}\\\\y'(\frac{\pi}{4})=\frac{2}{cos^20}=\frac{2}{1}=2\\\\2)\quad y=14\sqrt{2x-3}\; ,\; \; x_0=26\\\\y'=14\cdot \frac{1}{2\sqrt{2x-3}}\cdot 2=\frac{14}{\sqrt{2x-3}}\\\\y'(26)=\frac{14}{\sqrt{49}}=\frac{14}{7}=2\\\\3)\quad y= \frac{x-2}{x^2} \\\\y'= \frac{x^2-(x-2)\cdot 2x}{x^4} = \frac{-x^2+4x}{x^4} =\frac{4-x}{x^3}=0\\\\4-x=0\\\\x=4$