Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 206

\[1)\ y = \cos\left( 1 - \frac{x}{2} \right);\]

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

\[y^{'}(x) = \frac{1}{2}\sin\left( 1 - \frac{x}{2} \right).\]

\[2)\ y = \sin\left( 2 - \frac{3x}{4} \right);\]

\[y^{'}(x) = - \frac{3}{4}\cos\left( 2 - \frac{3x}{4} \right).\]

\[3)\ y = \sin\frac{x + 3}{2};\]

\[y^{'}(x) = \frac{1}{2}\cos\frac{x + 3}{2}.\]

\[4)\ y = \cos\frac{1 - x}{3};\]

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

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

\[5)\ y = \cos\frac{4 - 5x}{3};\]

\[y^{'}(x) = - \frac{5}{3} \bullet \left( - \sin\frac{4 - 5x}{3} \right);\]

\[y^{'}(x) = \frac{5}{3}\sin\frac{4 - 5x}{3}.\]

\[6)\ y = \sin\frac{2x + 3}{5};\]

\[y^{'}(x) = \frac{2}{5}\cos\frac{2x + 3}{5}.\]

\[7)\ y = \sin^{3}{2x};\]

\[y^{'}(x) = 2 \bullet 3\sin^{2}{2x} \bullet \cos{2x};\]

\[y^{'}(x) = 6\sin^{2}{2x} \bullet \cos{2x}.\]

\[8)\ y = \cos^{4}{3x};\]

\[y^{'}(x) = 3 \bullet 4\cos^{3}{3x} \bullet \left( - \sin{3x} \right);\]

\[y^{'}(x) = - 12\cos^{3}{3x} \bullet \sin{3x}.\]

\[9)\ y = ctg^{2}\ 4x;\]

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

\[y^{'}(x) = - \frac{8\cos{4x}}{\sin^{3}{4x}}.\]

\[10)\ y = tg^{4}\frac{x}{2};\]

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

\[y^{'}(x) = \frac{2\sin^{3}\frac{x}{2}}{\cos^{5}\frac{x}{2}}.\]