ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 201

\[1)\ y = (x + 3)^{8};\]

\[y^{'}(x) = 8(x + 3)^{7}.\]

\[2)\ y = (x - 4)^{7};\]

\[y^{'}(x) = 7(x - 4)^{6}.\]

\[3)\ y = \sqrt[3]{x - 2};\]

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

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

\[4)\ y = \sqrt{x + 5};\]

\[y^{'}(x) = \frac{1}{2}(x + 5)^{- \frac{1}{2}} =\]

\[= \frac{1}{2\sqrt{x + 5}}.\]

\[5)\ y = \frac{1}{(x + 1)^{2}};\]

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

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

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

\[y^{'}(x) = - 3 \bullet (x - 1)^{- 4} =\]

\[= - \frac{3}{(x - 1)^{4}}.\]

\[7)\ y = \frac{1}{\sqrt{x + 3}};\]

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

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

\[8)\ y = \frac{3}{\sqrt[3]{x - 4}};\]

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

\[= - \frac{1}{\sqrt[3]{(x - 4)^{4}}}.\]