Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1543

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\[1)\ f(x) = x^{3} - \frac{x^{2}}{2} + x;\text{\ \ }x_{0} = \frac{1}{3}:\]

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

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

\[= 3x^{2} - x + 1;\]

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

\[= 3 \bullet \frac{1}{9} - \frac{1}{3} + 1 = \frac{1}{3} - \frac{1}{3} + 1 = 1.\]

\[Ответ:\ \ 1.\]

\[2)\ f(x) = \frac{\ln x}{x};\text{\ \ }x_{0} = 1:\]

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

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

\[f^{'}(1) = \frac{1 - \ln 1}{1^{2}} = \frac{1 - 0}{1} = 1.\]

\[Ответ:\ \ 1.\]

\[3)\ f(x) = x^{- 3} - \frac{2}{x^{2}} + 3x;\ x_{0} = 3:\]

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

\[= - 3x^{- 4} - 2 \bullet \left( - 2x^{- 3} \right) + 3 =\]

\[= - \frac{3}{x^{4}} + \frac{4}{x^{3}} + 3;\]

\[f^{'}(3) = - \frac{3}{3^{4}} + \frac{4}{3^{3}} + 3 =\]

\[= - \frac{3}{81} + \frac{4}{27} + 3 = \frac{- 1 + 4}{27} + 3 =\]

\[= 3\frac{3}{27} = 3\frac{1}{9}.\]

\[Ответ:\ \ 3\frac{1}{9}.\]

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

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

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

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

\[= - \frac{\sin^{2}x + \cos^{2}x}{\sin^{2}x} = - \frac{1}{\sin^{2}x};\]

\[f^{'}\left( \frac{\pi}{4} \right) = - 1\ :\sin^{2}\frac{\pi}{4} =\]

\[= - 1\ :\left( \frac{1}{\sqrt{2}} \right)^{2} = - 1\ :\frac{1}{2} = - 2.\]

\[Ответ:\ \ - 2.\]