Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 210

\[1)\ y = \sqrt[3]{\frac{2x - 1}{3}} + \ln\frac{2x + 3}{5};\]

\[y^{'}(x) = \frac{2}{3} \bullet \frac{1}{3}\left( \frac{2x - 1}{3} \right)^{- \frac{2}{3}} + \frac{2}{5} \bullet \frac{5}{2x + 3} =\]

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

\[2)\ y = \sqrt{\frac{1 - x}{6}} - 2\ln\frac{2 - 5x}{3};\]

\[= - \frac{\sqrt{6}}{12\sqrt{1 - x}} + \frac{10}{2 - 5x}.\]

\[3)\ y = 2e^{\frac{1 - x}{3}} + 3\cos\frac{1 - x}{2};\]

\[= - \frac{2}{3}e^{\frac{1 - x}{3}} + \frac{3}{2}\sin\frac{1 - x}{2}.\]

\[4)\ y = 5\sin\frac{2x + 3}{4} - 4\sqrt{\frac{1}{x - 1}};\]

\[= \frac{5}{2}\cos\frac{2x + 3}{4} + \frac{2}{\sqrt{(x - 1)^{3}}}.\]

\[5)\ y = \sqrt[3]{\frac{1}{2 - x}} - 3\cos\frac{x - 2}{3};\]

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

\[6)\ y = 6\sqrt[3]{\frac{1}{(2 - x)^{2}}} + 4e^{\frac{3 - 5x}{2}};\]

\[= \frac{4}{\sqrt[3]{(2 - x)^{5}}} - 10e^{\frac{3 - 5x}{2}}.\]