Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 224

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

\[f^{'}(x) = 3x^{2};\]

\[f^{'}(1) = 3 \bullet 1^{2} = 3.\]

\[Ответ:\ \ 3.\]

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

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

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

\[Ответ:\ \ \frac{\sqrt{2}}{2}.\]

\[3)\ f(x) = \ln x,\ \ \ x_{0} = 1:\]

\[f^{'}(x) = \frac{1}{x};\]

\[f^{'}(1) = \frac{1}{1} = 1.\]

\[Ответ:\ \ 1.\]

\[4)\ f(x) = e^{x},\ \ \ x_{0} = \ln 3:\]

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

\[f^{'}\left( \ln 3 \right) = e^{\ln 3} = 3.\]

\[Ответ:\ \ 3.\]

\[5)\ f(x) = 3x^{2} - 4x,\ \ \ x_{0} = 2:\]

\[f^{'}(x) = 3 \bullet 2x - 4;\]

\[f^{'}(2) = 6 \bullet 2 - 4 = 8.\]

\[Ответ:\ \ 8.\]

\[6)\ f(x) = \sqrt{x} - \frac{1}{\sqrt{x}},\ \ \ x_{0} = 1:\]

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

\[f^{'}(1) = \frac{1}{2\sqrt{1}} + \frac{1}{2} \bullet 1^{- \frac{3}{2}} = 1.\]

\[Ответ:\ \ 1.\]