Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 835

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

\[f^{'}(x) = 2 \bullet \left( \ln x \right) + \left( 3^{x} \right)^{'} =\]

\[= 2 \bullet \frac{1}{x} + 3^{x} \bullet \ln 3 = \frac{2}{x} + 3^{x} \bullet \ln 3\]

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

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

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

\[3)\ f(x) = \log_{2}x + \frac{1}{2x}\]

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

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

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

\[4)\ f(x) = 3x^{- 3} - \log_{3}x\]

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

\[= 3 \bullet ( - 3) \bullet x^{- 4} - \frac{1}{x\ln 3} =\]

\[= - 9x^{- 4} - \frac{1}{x\ln 3}\]

\[5)\ f(x) = \ln\left( x^{2} - 2x \right)\]

\[u = x^{2} - 2xf(u) = \ln u:\]

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

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

\[6)\ f(x) = \left( 3x^{2} - 2 \right) \bullet \log_{3}x\]

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

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

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

\[= \frac{6x \bullet \ln x}{\ln 3} + \frac{3x^{2} - 2}{x\ln 3} =\]

\[= \frac{6x^{2} \bullet \ln x + 3x^{2} - 2}{x\ln 3} =\]

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

\[= \frac{3x \bullet \left( 2\ln x + 1 \right)}{\ln 3} - \frac{2}{x\ln 3}.\]