ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 246

\[1)\ y = \sin x\cos x + x;\]

\[y^{'}(x) =\]

\[= \cos x \bullet \cos x - \sin x \bullet \sin x + 1 =\]

\[= \cos^{2}x - \sin^{2}x + 1 = \cos{2x} + 1.\]

\[2)\ y = \left( x^{3} + 1 \right)\cos{2x};\]

\[y^{'}(x) =\]

\[= 3x^{2} \bullet \cos{2x} - 2\left( x^{3} + 1 \right) \bullet \sin{2x}.\]

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

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

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

\[= \frac{3x + 2x + 4}{3\sqrt[3]{x}} = \frac{5x + 4}{3\sqrt[3]{x}}.\]

\[4)\ y = \sqrt[3]{x - 1}\left( x^{4} - 1 \right);\]

\[y^{'}(x) =\]

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

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