ГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 68

Авторы:
Год:2020-2021-2022
Тип:учебник

Задание 68

\[\boxed{\mathbf{68}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = \sin x\]

\[f^{'}(x) = \cos x;\]

\[f^{''}(x) = - \sin x;\]

\[f^{'''}(x) = - \cos x;\]

\[f^{(4)}(x) = \sin x;\]

\[f^{(5)}(x) = \cos x;\]

\[f^{(6)}(x) = - \sin x;\]

\[f^{(7)}(x) = - \cos x;\]

\[f^{(8)}(x) = f^{(4)} = \sin x.\]

\[Получаем\ для\ любого\ n \in N:\]

\[f^{(4n - 3)}(x) = \cos{x;}\]

\[f^{(4n - 2)}(x) = - \sin x;\]

\[f^{(4n - 1)}(x) = - \cos x;\]

\[f^{(4n)}(x) = \sin x.\]

\[200 = 4 \cdot 50:\]

\[f^{(200)}(x) = \sin x.\]

\[\textbf{б)}\ f(x) = \cos x;\]

\[f^{'}(x) = - \sin x;\]

\[f^{''}(x) = - \cos x;\]

\[f^{'''}(x) = \sin x;\]

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

\[f^{(5)}(x) = - \sin x;\]

\[f^{(6)}(x) = - \cos x;\]

\[f^{(7)}(x) = \sin x;\]

\[f^{(8)}(x) = f^{(4)} = \cos x.\]

\[Получаем\ для\ любого\ n \in N:\]

\[f^{(4n - 3)}(x) = - \sin x\]

\[f^{(4n - 2)}(x) = - \cos x;\]

\[f^{(4n - 1)}(x) = \sin x;\]

\[f^{(4n)}(x) = \cos x.\]

\[200 = 4 \cdot 50:\]

\[f^{(200)}(x) = \cos x.\]

\[\textbf{в)}\ f(x) = e^{x};\]

\[f^{'}(x) = e^{x};\]

\[f^{(200)} = e^{x}.\]

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