ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 810

\[\boxed{\mathbf{810}\mathbf{.}}\]

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

\[= 5x^{4} - 4x^{3} + 3x^{2} - 2x.\]

\[2)\ f(x) = (x + 2) \bullet \sqrt[3]{x}\]

\[f^{'}(x) =\]

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

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

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

\[= \frac{4x + 2}{3\sqrt[3]{x^{2}}}.\]

\[3)\ f(x) = (x - 1) \bullet \sqrt{x}\]

\[f^{'}(x) =\]

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

\[= 1 \bullet \sqrt{x} + (x - 1) \bullet \frac{1}{2\sqrt{x}} =\]

\[= \sqrt{x} + \frac{x - 1}{2\sqrt{x}} = \frac{2x + x - 1}{2\sqrt{x}} =\]

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