ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 880

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

\[= \frac{4\sin{2x}}{4\cos^{4}x} = \frac{\sin{2x}}{\cos^{4}x}.\]

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

\[y^{'}(x) =\]

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

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

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

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

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

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

\[3)\ y = \frac{x}{\sqrt{x + 2}}\]

\[y^{'}(x) =\]

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

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

\[= \frac{2(x + 2) - x}{2\sqrt{x + 2}}\ :(x + 2) =\]

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

\[= \frac{x + 4}{2(x + 2)\sqrt{x + 2}}.\]

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

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

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