Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1127

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

\[= (8x + 4)\sqrt{x - 1} + \frac{(2x + 1)^{2}}{2\sqrt{x - 1}} =\]

\[= \frac{2(8x + 4)(x - 1) + (2x + 1)^{2}}{2\sqrt{x - 1}} =\]

\[= \frac{16x^{2} - 16x + 8x - 8 + 4x^{2} + 4x + 1}{2\sqrt{x - 1}} =\]

\[= \frac{20x^{2} - 4x - 7}{2\sqrt{x - 1}}.\]

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

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

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

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

\[= \frac{6x^{2} + 6x + 2x^{2}}{3\sqrt[3]{x + 1}} = \frac{8x^{2} + 6x}{3\sqrt[3]{x + 1}} =\]

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

\[3)\ y = \sin{2x} \bullet \cos{3x}\]

\[= \frac{5\cos(2x + 3x) - \cos(3x - 2x)}{2} =\]

\[= \frac{5\cos{5x} - \cos x}{2}.\]

\[4)\ y = x \bullet \cos{2x}\]

\[y^{'}(x) = (x)^{'} \bullet \cos{2x} + x \bullet \left( \cos{2x} \right)^{'} =\]

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

\[= \cos{2x} - 2x \bullet \sin{2x}.\]