ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 180

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

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

\[2)\ f(x) = \sqrt[5]{x} = x^{\frac{1}{5}};\]

\[f^{'}(x) = \frac{1}{5}x^{- \frac{4}{5}} = \frac{1}{5\sqrt[5]{x^{4}}}.\]

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

\[f^{'}(x) = 2 \bullet \frac{1}{6}x^{- \frac{5}{6}} - \frac{1}{3}x^{- \frac{2}{3}} =\]

\[= \frac{1}{3\sqrt[6]{x^{5}}} - \frac{1}{x\sqrt[3]{x^{2}}}.\]

\[4)\ f(x) = 3\sqrt[6]{x} + 7\sqrt[14]{x} =\]

\[= 3x^{\frac{1}{6}} + 7x^{\frac{1}{14}};\]

\[f^{'}(x) = 3 \bullet \frac{1}{6}x^{- \frac{5}{6}} + 7 \bullet \frac{1}{14}x^{- \frac{13}{14}} =\]

\[= \frac{1}{2\sqrt[6]{x^{5}}} + \frac{1}{2\sqrt[14]{x^{13}}}.\]

\[5)\ f(x) = \frac{2}{5\sqrt{x}} = \frac{2}{5}x^{- \frac{1}{2}};\]

\[f^{'}(x) = \frac{2}{5} \bullet \left( - \frac{1}{2}x^{- \frac{3}{2}} \right) =\]

\[= - \frac{1}{5\sqrt{x^{3}}} = - \frac{1}{5x\sqrt{x}}.\]

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

\[f^{'}(x) = - \frac{3}{2}x^{- \frac{5}{2}} = - \frac{3}{2\sqrt{x^{5}}} =\]

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

\[7)\ f(x) = \frac{x^{3} + 1}{x} = x^{2} + x^{- 1};\]

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

\[8)\ f(x) = \frac{x^{4} - \sqrt{x}}{x} = x^{3} - x^{- \frac{1}{2}};\]

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

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