ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 839

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\[1)\ f(x) = \frac{\cos x}{e^{x}}\]

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

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

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

\[f^{'}(x) = - \frac{\sin x + \cos x}{e^{x}}.\]

\[2)\ f(x) = \frac{3^{x}}{\sin x}\]

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

\[f^{'}(x) = \frac{3^{x} \bullet \ln 3 \bullet \sin x - 3^{x} \bullet \cos x}{\sin^{2}x}\]

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

\[3)\ f(x) = \ln x \bullet \cos{3x}\]

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

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

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

\[= \frac{\cos{3x}}{x} - 3\ln x \bullet \sin{3x}.\]

\[4)\ f(x) = \log_{3}x \bullet \sin{2x}\]

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

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

\[= \frac{1}{x\ln 3} \bullet \sin{2x} + \log_{3}x \bullet 2\cos{2x} =\]

\[= \frac{\sin{2x}}{x\ln 3} + 2\log_{3}x \bullet \cos{2x}.\]