Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 63

\[\boxed{\mathbf{63}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = \sqrt{x} = x^{\frac{1}{2}};\ \ x \geq 0\]

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

\[\ x > 0.\]

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

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

\[= \frac{1}{3} \cdot \frac{1}{\sqrt[3]{x^{2}}} = \frac{1}{3\sqrt[3]{x^{2}}};\ \ x \neq 0.\]

\[\textbf{в)}\ f(x) = \sqrt[3]{x^{2}} = x^{\frac{2}{3}};\ \ x \in R\]

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

\[= \frac{2}{3\sqrt[3]{x}};\ \ x \neq 0.\]

\[\textbf{г)}\ f(x) = \frac{1}{\sqrt{x}} = x^{- \frac{1}{2}};\ \ x > 0\]

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

\[= - \frac{1}{2\sqrt{x^{3}}};\ \ \ x > 0.\]

\[\textbf{д)}\ f(x) = \frac{1}{\sqrt[3]{x}} = x^{- \frac{1}{3}};\ \ x \neq 0\]

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

\[= - \frac{1}{\sqrt[3]{x^{4}}};\ \ x \neq 0.\]

\[\textbf{е)}\ f(x) = \frac{1}{\sqrt[3]{x^{2}}} = x^{- \frac{2}{3}};\ \ \ x \neq 0\]

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

\[= - \frac{2}{3\sqrt[3]{x^{5}}};\ \ x \neq 0.\]

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

\[= x^{\frac{5}{2}};\ \ x > 0\]

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

\[= 2,5x\sqrt{x};\ \ x > 0.\]

\[\textbf{з)}\ f(x) = \frac{x^{3}}{\sqrt[3]{x}} = x^{3} \cdot x^{- \frac{1}{3}} = x^{\frac{8}{3}};\ \]

\[\ \ x \neq 0\]

\[f^{'}(x) = \frac{8}{3} \cdot x^{\frac{5}{3}} = \frac{8}{3} \cdot \sqrt[3]{x^{5}} =\]

\[= 2\frac{2}{3} \cdot x\sqrt[3]{x^{2}};\ \ x \neq 0.\]