ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 252

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

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

\[2)\ y = \sqrt{\ln x};\]

\[y^{'}(x) = \frac{1}{x} \bullet \frac{1}{2\sqrt{\ln x}} = \frac{1}{2x\sqrt{\ln x}}.\]

\[3)\ y = \sin\sqrt{x};\]

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

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

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

\[= - 4\cos^{3}x \bullet \sin x.\]

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

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

\[= \frac{1}{2\cos^{2}x\sqrt{\text{tg\ x}}}.\]

\[6)\ y = ctg\ 3x;\]

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

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