ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 253

\[1)\ y = \cos^{2}{3x};\]

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

\[= - 6\sin{3x}\cos{3x} = - 3\sin{6x}.\]

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

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

\[= \frac{\sin\frac{x}{2}}{\cos^{3}\frac{x}{2}}.\]

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

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

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

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

\[4)\ y = \cos\left( x + 2x^{3} \right);\]

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

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

\[= - \left( 1 + 6x^{2} \right)\sin\left( x + 2x^{3} \right).\]

\[5)\ y = e^{\text{tg\ x}};\]

\[y^{'}(x) = \frac{1}{\cos^{2}x} \bullet e^{\text{tg\ x}} = \frac{e^{\text{tg\ x}}}{\cos^{2}x}.\]

\[6)\ y = \cos\left( e^{x} \right);\]

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

\[= - e^{x} \bullet \sin e^{x}.\]

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

\[y^{'}(x) = 2x \bullet 3^{x^{2}} \bullet \ln 3.\]

\[8)\ y = 2^{\cos x};\]

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