Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 840

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\[1)\ f(x) = e^{2x - 4} + 2\ln x\ x_{0} = 2:\]

\[f^{'}(x) = \left( e^{2x - 4} \right)^{'} + 2 \bullet \left( \ln x \right)^{'} =\]

\[= 2e^{2x - 4} + 2 \bullet \frac{1}{x}\]

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

\[= 2e^{0} + 1 = 2 \bullet 1 + 1 = 3.\]

\[Ответ:\ \ 3.\]

\[2)\ f(x) = e^{3x - 2} - \ln(3x - 1)\]

\[x_{0} = \frac{2}{3}:\]

\[f^{'}(x) = \left( e^{3x - 2} \right)^{'} - \left( \ln(3x - 1) \right)^{'} =\]

\[= 3e^{3x - 2} - \frac{3}{3x - 1}\]

\[f^{'}\left( \frac{2}{3} \right) = 3e^{3 \bullet \frac{2}{3} - 2} - \frac{3}{3 \bullet \frac{2}{3} - 1} =\]

\[= 3e^{0} - \frac{3}{2 - 1} = 3 \bullet 1 - 3 = 0.\]

\[Ответ:\ \ 0.\]

\[3)\ f(x) = 2^{x} - \log_{2}x\ x_{0} = 1:\]

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

\[= 2^{x} \bullet \ln 2 - \frac{1}{x\ln 2}\]

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

\[= 2\ln 2 - \frac{1}{\ln 2} = \frac{2\ln^{2}2 - 1}{\ln 2}.\]

\[Ответ:\ \ \frac{2\ln^{2}2 - 1}{\ln 2}.\]

\[4)\ f(x) = \log_{0,5}x - 3^{x}\ x_{0} = 1:\]

\[f^{'}(x) = \left( \log_{0,5}x \right)^{'} - \left( 3^{x} \right)^{'} =\]

\[= \frac{1}{x\ln{0,5}} - 3^{x} \bullet \ln 3\]

\[f^{'}(1) = \frac{1}{1 \bullet \ln{0,5}} - 3^{1} \bullet \ln 3 =\]

\[= - \frac{1}{\ln 2} - 3\ln 3.\]

\[Ответ:\ - \frac{1}{\ln 2} - 3\ln 3.\]