Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1556

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\[f(x) = \frac{1 + \sin{2x}}{1 - \sin{2x}};\]

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

\[f^{'}(0) = \frac{4\cos(2 \bullet 0)}{\left( 1 - \sin(2 \bullet 0) \right)^{2}} =\]

\[= \frac{4\cos 0}{\left( 1 - \sin 0 \right)^{2}} = \frac{4 \bullet 1}{(1 - 0)^{2}} =\]

\[= \frac{4}{1^{2}} = 4;\]

\[f^{'}\left( \frac{\pi}{6} \right) = \frac{4\cos\frac{2\pi}{6}}{\left( 1 - \sin\frac{2\pi}{6} \right)^{2}} =\]

\[= \frac{4\cos\frac{\pi}{3}}{\left( 1 - \sin\frac{\pi}{3} \right)^{2}} = \frac{4 \bullet \frac{1}{2}}{\left( 1 - \frac{\sqrt{3}}{2} \right)^{2}} =\]

\[= \frac{2}{\left( \frac{2 - \sqrt{3}}{2} \right)^{2}} = \frac{8}{\left( 2 - \sqrt{3} \right)^{2}} =\]

\[= \frac{8}{4 - 4\sqrt{3} + 3} = \frac{8}{7 - 4\sqrt{3}} =\]

\[= \frac{8\left( 7 + 4\sqrt{3} \right)}{49 - 48} = 8\left( 7 + 4\sqrt{3} \right).\]

\[Ответ:\ \ f^{'}(0) = 4;\ \ \]

\[f^{'}\left( \frac{\pi}{6} \right) = 8\left( 7 + 4\sqrt{3} \right)\text{.\ }\]