ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 789

\[\boxed{\mathbf{789}\mathbf{.}}\]

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

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

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

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

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

\[3)\ f(x) = x^{- \frac{2}{7}}\]

\[f^{'}(x) = \left( x^{- \frac{2}{7}} \right) = - \frac{2}{7} \bullet x^{- \frac{2}{7} - 1} =\]

\[= - \frac{2}{7} \bullet x^{- \frac{9}{7}} = - \frac{2}{7} \bullet \frac{1}{x^{\frac{9}{7}}} = - \frac{2}{7\sqrt[7]{x^{9}}}\]

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

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