Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 62

\[\boxed{\mathbf{62}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = x^{0,5} = \sqrt{x};\ \ \ x \geq 0\]

\[f^{'}(x) = 0,5x^{0,5 - 1} = 0,5x^{- 0,5} =\]

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

\[\textbf{б)}\ f(x) = x^{- 0,5} = \frac{1}{\sqrt{x}};\ \ x > 0\]

\[f^{'}(x) = - 0,5x^{- 1,5} = - \frac{1}{2\sqrt{x^{3}}};\ \]

\[\ x > 0.\]

\[\textbf{в)}\ f(x) = x^{4,2} = x^{\frac{21}{5}};\ \ x \in R\]

\[f^{'}(x) = 4,2x^{3,2};\ \ x \in R.\]

\[\textbf{г)}\ f(x) = x^{- 0,2} = \frac{1}{\sqrt[5]{x}};\ \ x \neq 0\]

\[f^{'}(x) = - 0,2x^{- 1,2};\ \ x \neq 0.\]

\[\textbf{д)}\ f(x) = x^{\frac{4}{3}};\ \ x \in R\]

\[f^{'}(x) = \frac{4}{3}x^{\frac{1}{3}} = \frac{4\sqrt[3]{x}}{3};\ \ x \in R.\]

\[\textbf{е)}\ f(x) = x^{\frac{13}{3}};\ \ x \in R\]

\[f^{'}(x) = \frac{13}{3}x^{\frac{10}{3}} = \frac{13\sqrt[3]{x^{10}}}{3};\ \ x \in R.\]

\[\textbf{ж)}\ f(x) = x^{- 3,5} = \frac{1}{x^{3,5}};\ \ \ x > 0\]

\[f^{'}(x) = - 3,5x^{- 4,5};\ \ \ x > 0.\]

\[\textbf{з)}\ f(x) = x^{\frac{16}{3}};\ \ \ x \in R\]

\[f^{'}(x) = \frac{16}{3}x^{\frac{13}{3}} = \frac{16\sqrt[3]{x^{13}}}{3};\ \ x \in R.\]