ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 790

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\[1)\ f(x) = \frac{1}{x^{5}} = x^{- 5}\]

\[f^{'}(x) = - 5 \bullet x^{- 5 - 1} =\]

\[= - 5 \bullet x^{- 6} = - \frac{5}{x^{6}}\]

\[2)\ f(x) = \frac{1}{x^{9}} = x^{- 9}\]

\[f^{'}(x) = - 9 \bullet x^{- 9 - 1} =\]

\[= - 9 \bullet x^{- 10} = - \frac{9}{x^{10}}\]

\[3)\ f(x) = \sqrt[4]{x} = x^{\frac{1}{4}}\]

\[f^{'}(x) = \frac{1}{4} \bullet x^{\frac{1}{4} - 1} = \frac{1}{4} \bullet x^{- \frac{3}{4}} =\]

\[= \frac{1}{4} \bullet \frac{1}{x^{\frac{3}{4}}} = \frac{1}{4\sqrt[4]{x^{3}}}\]

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

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

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

\[5)\ f(x) = \frac{1}{\sqrt[3]{x}} = \frac{1}{x^{\frac{1}{3}}} = x^{- \frac{1}{3}}\]

\[f^{'}(x) = - \frac{1}{3} \bullet x^{- \frac{1}{3} - 1} = - \frac{1}{3} \bullet x^{- \frac{4}{3}} =\]

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

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

\[f^{'}(x) = - \frac{3}{4} \bullet x^{- \frac{3}{4} - 1} = - \frac{3}{4} \bullet x^{- \frac{7}{4}} =\]

\[= - \frac{3}{4} \bullet \frac{1}{x \bullet x^{\frac{3}{4}}} = - \frac{3}{4\sqrt[4]{x^{3}}}.\]