ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 200

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

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

\[= 6x - \frac{4}{3\sqrt[3]{x^{2}}} + \frac{2}{3}e^{\frac{x}{3}}.\]

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

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

\[= 6x^{2} + \frac{3}{2\sqrt{x}} + 2\sin{2x}.\]

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

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

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

\[4)\ y = 2x^{8} - 3\ tg\ 3x - \frac{1}{3}\sin{3x};\]

\[y^{'}(x) = 2 \bullet 8x^{7} - 3 \bullet \frac{3}{\cos^{2}{3x}} - \frac{1}{3} \bullet 3\cos{3x} =\]

\[= 16x^{7} - \frac{9}{\cos^{2}{3x}} - \cos{3x}.\]

\[5)\ y = 8x^{\frac{3}{4}} + 7x^{\frac{1}{7}} - \cos{4x};\]

\[y^{'}(x) = 8 \bullet \frac{3}{4}x^{- \frac{1}{4}} + 7 \bullet \frac{1}{7}x^{- \frac{6}{7}} + 4\sin{4x} =\]

\[= \frac{6}{\sqrt[4]{x}} + \frac{1}{\sqrt[7]{x^{6}}} + 4\sin{4x}.\]

\[6)\ y = \frac{1}{5}ctg\ x - 5x^{\frac{4}{5}} - \frac{1}{4}e^{2x};\]

\[y^{'}(x) = \frac{1}{5} \bullet \left( - \frac{1}{\sin^{2}x} \right) - 5 \bullet \frac{4}{5}x^{- \frac{1}{5}} - \frac{1}{4} \bullet 2e^{2x} =\]

\[= - \frac{1}{5\sin^{2}x} - \frac{4}{\sqrt[5]{x}} - \frac{1}{2}e^{2x}.\]