ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 237

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

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

\[= 8x^{3} - 3x^{2} + 3.\]

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

\[y^{'}(x) = - 5x^{4} + 2 \bullet 3x^{2} - 3 \bullet 2x =\]

\[= - 5x^{4} + 6x^{2} - 6x.\]

\[3)\ y = 6\sqrt[3]{x} + \frac{1}{x^{2}};\]

\[y^{'}(x) = 6 \bullet \frac{1}{3}x^{- \frac{2}{3}} - 2 \bullet x^{- 3} =\]

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

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

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

\[y^{'}(x) = - \frac{6}{x^{4}} - \frac{2}{\sqrt[4]{x^{3}}}.\]

\[5)\ y = (2x + 3)^{8};\]

\[y^{'}(x) = 2 \bullet 8(2x + 3)^{7} =\]

\[= 16(2x + 3)^{7}.\]

\[6)\ y = (4 - 3x)^{7};\]

\[y^{'}(x) = - 3 \bullet 7(4 - 3x)^{6} =\]

\[= - 21(4 - 3x)^{6}.\]

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

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

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

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

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

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

\[9)\ y = \sin{0,5x};\]

\[y^{'}(x) = 0,5\cos{0,5x}.\]

\[10)\ y = \cos( - 3x);\]

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

\[= - 3\sin{3x}.\]