Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 807

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

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

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

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

\[= - \frac{3}{27} - \frac{2}{27} = - \frac{5}{27}\]

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

\[= - 1 - 2 = - 3.\]

\[2)\ f(x) = \sqrt{x} + \frac{1}{x} + 1\]

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

\[= \frac{1}{2\sqrt{x}} - \frac{1}{x^{2}}\]

\[f^{'}(3) = \frac{1}{2\sqrt{3}} - \frac{1}{3^{2}} = \frac{\sqrt{3}}{2 \bullet 3} - \frac{1}{9} =\]

\[= \frac{3\sqrt{3} - 2}{18}\]

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

\[= - \frac{1}{2} = - 0,5.\]

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

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

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

\[= 3 \bullet \left( - \frac{1}{2} \right) \bullet x^{- \frac{3}{2}} - 2 \bullet ( - 3) \bullet x^{- 4} =\]

\[= \frac{6}{x^{4}} - \frac{3}{2\sqrt{x^{3}}}\]

\[f^{'}(3) = \frac{6}{3^{4}} - \frac{3}{2\sqrt{3^{3}}} = \frac{6}{81} - \frac{3}{2\sqrt{27}} =\]

\[= \frac{2}{27} - \frac{3}{2\sqrt{27}} = \frac{4 - 3\sqrt{27}}{54}\]

\[f^{'}(1) = \frac{6}{1^{4}} - \frac{3}{2\sqrt{1^{3}}} = 6 - \frac{3}{2} =\]

\[= 6 - 1,5 = 4,5.\]

\[4)\ f(x) = x^{\frac{3}{2}} - x^{- \frac{3}{2}}\]

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

\[= \frac{3}{2} \bullet x^{\frac{1}{2}} + \frac{3}{2} \bullet x^{- \frac{5}{2}} = \frac{3\sqrt{x}}{2} + \frac{3}{2\sqrt{x^{5}}} =\]

\[= \frac{3\sqrt{x}}{2} + \frac{3}{2x^{2}\sqrt{x}}\]

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

\[= \frac{3\sqrt{3}}{2} + \frac{3}{18\sqrt{3}} =\]

\[= \frac{81\sqrt{3} + 3\sqrt{3}}{54} = \frac{14\sqrt{3}}{9}\]

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

\[= \frac{3}{2} + \frac{3}{2} = 3.\]