Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 65

\[\boxed{\mathbf{65}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = 4\sin x \cdot \cos x\]

\[f^{'}(x) = \left( 4\sin x \right)^{'} \cdot \cos x +\]

\[+ 4\sin x \cdot \left( \cos x \right)^{'} = 4\cos x \cdot\]

\[\cdot \cos x - 4\sin x \cdot \sin x =\]

\[= 4 \cdot \left( \cos^{2}x - \sin^{2}x \right) = 4\cos{2x};\]

\[\ \ x \in R.\]

\[\textbf{б)}\ f(x) = \cos^{2}{3x} - \sin^{2}{3x} =\]

\[= \cos{6x};\ \ x \in R\]

\[f^{'}(x) = - 6\sin{6x};\ \ x \in R.\]

\[\textbf{в)}\ f(x) = \frac{2tg(1000x)}{1 - tg^{2}(1000x)} =\]

\[= tg(2000x);\ \ \]

\[x \neq \frac{\pi}{4000} + \frac{\text{πk}}{2000};\]

\[f^{'}(x) = \frac{2000}{\cos^{2}{2000x}};\ \ \]

\[x \neq \frac{\pi}{4000} + \frac{\text{πk}}{2000}.\]

\[\textbf{г)}\ f(x) = \sqrt[17]{\sin^{2}{7x} + \cos^{2}{7x}} =\]

\[= \sqrt[17]{1} = 1;\ \ x \in R\]

\[f^{'}(x) = 0;\ \ x \in R.\]