Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 243

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

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

\[= - \sin x \bullet \sin x + \cos x \bullet \cos x;\]

\[f^{'}(x) = \cos^{2}x - \sin^{2}x = \cos{2x};\]

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

\[Ответ:\ \ \frac{1}{2}.\]

\[2)\ f(x) = e^{x}\ln x,\ \ \ x_{0} = 1;\]

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

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

\[f^{'}(1) = 1 \bullet 0 + e \bullet 1 = 1;\]

\[Ответ:\ \ e.\]

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

\[f^{'}(x) = (2\ ctg\ x)^{'} =\]

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

\[f^{'}\left( \frac{\pi}{4} \right) = - 2\ :\left( \frac{\sqrt{2}}{2} \right)^{2} =\]

\[= - 2 \bullet 2 = - 4.\]

\[Ответ:\ - 4.\]

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

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

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

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

\[Ответ:\ \ \frac{1}{2}.\]