Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 889

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\[1)\ y = 2\sin\frac{x}{2}\text{\ \ }x_{0} = \frac{3\pi}{2}:\]

\[y^{'}(x) = 2 \bullet \left( \sin\frac{x}{2} \right)^{'} =\]

\[= 2 \bullet \frac{1}{2} \bullet \cos\frac{x}{2} = \cos\frac{x}{2}\]

\[y^{'}\left( \frac{3\pi}{2} \right) = \cos\frac{3\pi}{4} =\]

\[= \cos\left( \pi - \frac{\pi}{4} \right) = - \cos\frac{\pi}{4} = - \frac{\sqrt{2}}{2}\]

\[y\left( \frac{3\pi}{2} \right) = 2 \bullet \sin\frac{3\pi}{4} =\]

\[= 2 \bullet \sin\left( \pi - \frac{\pi}{4} \right) = 2 \bullet \sin\frac{\pi}{4} =\]

\[= 2 \bullet \frac{\sqrt{2}}{2} = \sqrt{2}\]

\[y = \sqrt{2} - \frac{\sqrt{2}}{2} \bullet \left( x - \frac{3\pi}{2} \right) =\]

\[= - \frac{\sqrt{2}}{2}x + \sqrt{2} + \frac{3\pi\sqrt{2}}{4}.\]

\[Ответ:\ \ \]

\[y = - \frac{\sqrt{2}}{2}x + \sqrt{2} + \frac{3\pi\sqrt{2}}{4}.\]

\[2)\ y = 2^{- x} - 2^{- 2x}\text{\ \ }x_{0} = 2:\]

\[y^{'} = \left( 2^{- x} \right)^{'} - \left( 2^{- 2x} \right)^{'} =\]

\[= - 2^{- x} \bullet \ln 2 + 2 \bullet 2^{- 2x} \bullet \ln 2 =\]

\[= \ln 2 \bullet \left( 2^{1 - 2x} - 2^{- x} \right)\]

\[y^{'}(2) = \ln 2 \bullet \left( 2^{1 - 2 \bullet 2} - 2^{- 2} \right) =\]

\[= \ln 2 \bullet \left( 2^{- 3} - 2^{- 2} \right) =\]

\[= \ln 2 \bullet \left( \frac{1}{8} - \frac{1}{4} \right) = - \frac{\ln 2}{8}\]

\[y(2) = 2^{- 2} - 2^{- 2 \bullet 2} =\]

\[= 2^{- 2} - 2^{- 4} = \frac{1}{4} - \frac{1}{16} = \frac{3}{16}\]

\[y = \frac{3}{16} - \frac{\ln 2}{8} \bullet (x - 2) =\]

\[= - \frac{\ln 2}{8}x + \frac{3}{16} + \frac{\ln 2}{4}.\]

\[Ответ:\ \ y = - \frac{\ln 2}{8}x + \frac{3}{16} + \frac{\ln 2}{4}.\]

\[3)\ y = \frac{x + 2}{3 - x}\text{\ \ }x_{0} = 2:\]

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

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

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

\[= \frac{3 - x + x + 2}{(3 - x)^{2}} = \frac{5}{(3 - x)^{2}}\]

\[y^{'}(2) = \frac{5}{(3 - 2)^{2}} = \frac{5}{1^{2}} = 5\]

\[y(2) = \frac{2 + 2}{3 - 2} = \frac{4}{1} = 4\]

\[y = 4 + 5(x - 2) =\]

\[= 4 + 5x - 10 = 5x - 6.\]

\[Ответ:\ \ y = 5x - 6.\]

\[4)\ y = x + \ln x\text{\ \ }x_{0} = e:\]

\[y^{'}(x) = (x)^{'} + \left( \ln x \right)^{'} = 1 + \frac{1}{x}\]

\[y^{'}(e) = 1 + \frac{1}{e}\]

\[y(e) = e + \ln e = e + 1\]

\[y = e + 1 + \left( 1 + \frac{1}{e} \right) \bullet (x - e) =\]

\[= e + 1 + x - e + \frac{x}{e} - 1 = x + \frac{x}{e}.\]

\[Ответ:\ \ y = x + \frac{x}{e}.\]