Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 862

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\[1)\ f(x) = x + \frac{1}{x + 1}\text{\ \ }и\ \ x_{0} = 0:\]

\[f^{'}(x) = (x)^{'} + {(x + 1)^{- 1}}^{'} =\]

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

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

\[f^{'}(0) = 1 - \frac{1}{(0 + 1)^{2}} =\]

\[= 1 - \frac{1}{1^{2}} = 1 - 1 = 0\]

\[f(0) = 0 + \frac{1}{0 + 1} = \frac{1}{1} = 1\]

\[y = 1 + 0(x - 0) = 1.\]

\[Ответ:\ \ y = 1.\]

\[2)\ f(x) = \sin{2x} - \ln(x + 1)\ и\ \ \]

\[x_{0} = 0:\]

\[f^{'}(x) = \left( \sin{2x} \right)^{'} - \left( \ln(x + 1) \right)^{'} =\]

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

\[f^{'}(0) = 2\cos(2 \bullet 0) - \frac{1}{0 + 1} =\]

\[= 2 \bullet \cos 0 - \frac{1}{1} = 2 \bullet 1 - 1 = 1\]

\[f(0) = \sin(2 \bullet 0) - \ln(0 + 1) =\]

\[= \sin 0 - \ln 1 = 0 - 0 = 0\]

\[y = 0 + 1(x - 0) = x.\]

\[Ответ:\ \ y = x.\]