Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 882

\[\boxed{\mathbf{882}\mathbf{.}}\]

\[\textbf{а)}\ y = e^{- x}\]

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

\[y^{'}(0) = - e^{0} = - 1\]

\[Ответ:\ \ г.\]

\[\textbf{б)}\ y = \ln( - x)\]

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

\[y^{'}( - 1) = \frac{1}{- 1} = - 1\]

\[Ответ:\ \ а.\]

\[\textbf{в)}\ y = \sin{2x}\]

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

\[2\cos{2x} = 0\]

\[\cos{2x} = 0\]

\[2x = \arccos 0 + \pi n\]

\[2x = \frac{\pi}{2} + \pi n\]

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ в.\]

\[\textbf{г)}\ y = 2\cos x\]

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

\[- 2\sin x = 0\]

\[\sin x = 0\]

\[x = \arcsin 0 + \pi n = \pi n.\]

\[Ответ:\ \ б.\]