Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 876

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\[1)\ f(x) = \cos x \bullet \sin x\text{\ \ }x_{0} = \frac{\pi}{6}:\]

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

\[= \left( \cos x \right)^{'} \bullet \sin x + \cos x \bullet \left( \sin x \right)^{'} =\]

\[= - \sin^{2}x + \cos^{2}x = \cos{2x}\]

\[f^{'}\left( \frac{\pi}{6} \right) = \cos\left( 2 \bullet \frac{\pi}{6} \right) = \cos\frac{\pi}{3} = \frac{1}{2}.\]

\[Ответ:\ \ \frac{1}{2}.\]

\[2)\ f(x) = e^{x} \bullet \ln x\text{\ \ }x_{0} = 1:\]

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

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

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

\[f^{'}(1) = e^{1} \bullet \left( \ln 1 + \frac{1}{1} \right) =\]

\[= e \bullet (0 + 1) = e.\]

\[Ответ:\ \ e.\]

\[3)\ f(x) = \frac{2\cos x}{\sin x}\text{\ \ }x_{0} = \frac{\pi}{4}:\]

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

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

\[= \frac{- 2\sin x \bullet \sin x - 2\cos x \bullet \cos x}{\sin^{2}x} =\]

\[= - 2 \bullet \frac{\sin^{2}x + \cos^{2}x}{\sin^{2}x} =\]

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

\[f^{'}\left( \frac{\pi}{4} \right) = - 2 \bullet \left( 1 + ctg^{2}\frac{\pi}{4} \right) =\]

\[= - 2 \bullet \left( 1 + 1^{2} \right) = - 2 \bullet 2 = - 4.\]

\[Ответ:\ \ - 4.\]

\[4)\ f(x) = \frac{x}{1 + e^{x}}\text{\ \ }x_{0} = 0:\]

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

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

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

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

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

\[Ответ:\ \ \frac{1}{2}.\]