Страница 258 номер 880, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(y = \frac{1 - \cos 2x}{1 + \cos 2x}\)
\(y'(x) = \dfrac{2 \sin 2x(1 + \cos 2x) + 2(1 - \cos 2x) \sin 2x}{(1 + \cos 2x)^{2}} =\)
\(= \dfrac{2 \sin 2x + 2 \sin 2x \cos 2x + 2 \sin 2x - 2 \cos 2x \sin 2x}{(1 + \cos 2x)^{2}} =\)
\(= \dfrac{4 \sin 2x}{(2 \cos^{2} x)^{2}} = \dfrac{\sin 2x}{\cos^{4} x};\)

2) \(y = \frac{\sqrt{x + 4}}{4x}\)
\(y'(x) = \dfrac{\frac{4x}{2\sqrt{x + 4}} - 4\sqrt{x + 4}}{16x^{2}} = \dfrac{4x - 8x - 32}{32x^{2}\sqrt{x + 4}} = \dfrac{-x - 8}{8x^{2}\sqrt{x + 4}};\)

3) \(y = \frac{x}{\sqrt{x + 2}}\)
\(y'(x) = \dfrac{\sqrt{x + 2} - \frac{x}{2\sqrt{x + 2}}}{x + 2} = \dfrac{2x + 4 - x}{(2x + 4)\sqrt{x + 2}} = \dfrac{x + 4}{(2x + 4)\sqrt{x + 2}};\)

4) \(y = \frac{\sin x + \cos x}{\sin x - \cos x}\)
\(y'(x) = \dfrac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^{2}} =\)
\(= \dfrac{2 \sin x \cos x - \sin^{2} x - \cos^{2} x - \sin^{2} x - 2 \sin x \cos x - \cos^{2} x}{1 - \sin 2x} =\)
\(= \dfrac{-2}{1 - \sin 2x}.\)