ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 203

\[1)\ y = \sqrt[4]{2 - 8x};\]

\[y^{'}(x) = - 8 \bullet \frac{1}{4} \bullet (2 - 8x)^{- \frac{3}{4}} =\]

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

\[2)\ y = \sqrt[3]{4x + 1};\]

\[y^{'}(x) = 4 \bullet \frac{1}{3} \bullet (4x + 1)^{- \frac{2}{3}} =\]

\[= \frac{4}{3\sqrt[3]{(4x + 1)^{2}}}.\]

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

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

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

\[4)\ y = \frac{1}{\sqrt{4x + 1}};\]

\[y^{'}(x) = 4 \bullet \left( - \frac{1}{2} \right) \bullet (4x + 1)^{- \frac{3}{2}} =\]

\[= - \frac{2}{\sqrt{(4x + 1)^{3}}}.\]

\[5)\ y = \frac{7}{\sqrt[4]{3 - 8x}};\]

\[y^{'}(x) = 7 \bullet ( - 8) \bullet \left( - \frac{1}{4} \right) \bullet (3 - 8x)^{- \frac{5}{4}} =\]

\[= \frac{14}{\sqrt[4]{(3 - 8x)^{5}}}.\]

\[6)\ y = \frac{7}{\sqrt[3]{2 - 9x}};\]

\[y^{'}(x) = 7 \bullet ( - 9) \bullet \left( - \frac{1}{3} \right) \bullet (2 - 9x)^{- \frac{4}{3}} =\]

\[= \frac{21}{\sqrt[3]{(2 - 9x)^{4}}}.\]