880 Найти производную функции 1) y = 1- cos 2x 1 + cos 2x x 3) y= √x+2 √x +4 -; 4x 4) y = sin x + cos x sin x - cos x

Найдем производную функции \( y = \frac{1 - \cos 2x}{1 + \cos 2x} \):

\[ y = \frac{1 - \cos 2x}{1 + \cos 2x} = \frac{2 \sin^2 x}{2 \cos^2 x} = \tan^2 x \]

\[y' = (\tan^2 x)' = 2 \tan x \cdot (\tan x)' = 2 \tan x \cdot \frac{1}{\cos^2 x} = 2 \tan x (1 + \tan^2 x)\]

Найдем производную функции \( y = \frac{\sqrt{x + 4}}{4x} \):

\[ y' = \frac{(\sqrt{x + 4})' \cdot 4x - \sqrt{x + 4} \cdot (4x)'}{(4x)^2} = \frac{\frac{1}{2\sqrt{x + 4}} \cdot 4x - 4\sqrt{x + 4}}{16x^2} = \frac{\frac{2x}{\sqrt{x + 4}} - 4\sqrt{x + 4}}{16x^2} = \frac{2x - 4(x + 4)}{16x^2 \sqrt{x + 4}} = \frac{-2x - 16}{16x^2 \sqrt{x + 4}} = \frac{-x - 8}{8x^2 \sqrt{x + 4}} \]

Найдем производную функции \( y = \frac{x}{\sqrt{x + 2}} \):

\[y' = \frac{x' \sqrt{x + 2} - x(\sqrt{x + 2})'}{(\sqrt{x + 2})^2} = \frac{\sqrt{x + 2} - x \cdot \frac{1}{2\sqrt{x + 2}}}{x + 2} = \frac{\frac{2(x + 2) - x}{2\sqrt{x + 2}}}{x + 2} = \frac{x + 4}{2(x + 2)\sqrt{x + 2}}\]

Найдем производную функции \( y = \frac{\sin x + \cos x}{\sin x - \cos x} \):

\[y' = \frac{(\sin x + \cos x)'(\sin x - \cos x) - (\sin x + \cos x)(\sin x - \cos x)'}{(\sin x - \cos x)^2} = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2} = \frac{-(\cos x - \sin x)^2 - (\sin x + \cos x)^2}{(\sin x - \cos x)^2} = \frac{-(\cos^2 x - 2 \sin x \cos x + \sin^2 x) - (\sin^2 x + 2 \sin x \cos x + \cos^2 x)}{(\sin x - \cos x)^2} = \frac{-2(\sin^2 x + \cos^2 x)}{(\sin x - \cos x)^2} = \frac{-2}{(\sin x - \cos x)^2}\]