Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 247

\[1)\ y = \frac{1 - \cos{2x}}{1 + \cos{2x}};\]

\[= \frac{4\sin{2x}}{\left( 1 + \cos{2x} \right)^{2}}.\]

\[2)\ y = \frac{\sqrt{x + 4}}{4x};\]

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

\[= \frac{2x - (x + 4) \bullet 4}{\sqrt{x + 4} \bullet 16x^{2}} =\]

\[= \frac{x - 2x - 8}{8x^{2} \bullet \sqrt{x + 4}} = - \frac{x + 8}{8x^{2} \bullet \sqrt{x + 4}}.\]

\[3)\ y = \frac{x}{\sqrt{x + 3}};\]

\[y^{'}(x) = \frac{\sqrt{x + 3} - x \bullet \frac{1}{2\sqrt{x + 3}}}{x + 3} =\]

\[= \frac{2(x + 3) - x}{2\sqrt{(x + 3)^{3}}} = \frac{x + 6}{2\sqrt{(x + 3)^{3}}}.\]

\[4)\ y = \frac{\sin x + \cos x}{\sin x - \cos x};\]

\[= \frac{- 1 - 1}{1 - \sin{2x}} = \frac{2}{\sin{2x} - 1}.\]