ГДЗ по алгебре 11 класс Никольский Параграф 4. Производная Задание 49

\[\boxed{\mathbf{49}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = \cos{2002x}\cos{2001x} +\]

\[+ \sin{2001x}\sin{2002x} =\]

\[= \cos(2002x - 2001x) = \cos x;\]

\[\ \ x \in R\]

\[f(x) = - \sin x.\]

\[\textbf{б)}\ f(x) = \sin{2002x}\cos{2001x} -\]

\[- \sin{2001x}\cos{2002x} =\]

\[= \sin(2002x - 2001x) = \sin x;\]

\[\ \ x \in R\]

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

\[\textbf{в)}\ f(x) =\]

\[= \frac{tg\ 2002x - tg\ 2001x}{1 + tg\ 2002x \cdot tg\ 2001x} =\]

\[= tg(2002x - 2001x) = tgx;\ \ \ \ \]

\[x \neq \frac{\pi}{2} + \pi k;\ \ k \in Z\]

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