Упр.28.33 ГДЗ Мордкович 10-11 класс (Алгебра)
а) у = sin (3x — пи/4), x0 = пи/4;
б) у = tg 6x, x0 = пи/24;
в) у = cos (пи/3 — 2x), x0 = пи/3;
г) у = ctg x/3, x0 = пи.
$$y=\sin\left(3x-\frac{\pi}{4}\right), \quad x_0=\frac{\pi}{4}$$
$$y’ = 3\cos\left(3x-\frac{\pi}{4}\right)$$
$$y’\left(\frac{\pi}{4}\right)=3\cos\left(3\cdot\frac{\pi}{4}-\frac{\pi}{4}\right)=3\cos\frac{\pi}{2}=0$$
$$y=\tg 6x, \quad x_0=\frac{\pi}{24}$$
$$y’=\frac{6}{\cos^2 6x}$$
$$y’\left(\frac{\pi}{24}\right)=\frac{6}{\cos^2\left(6\cdot\frac{\pi}{24}\right)}=\frac{6}{\cos^2\frac{\pi}{4}}=\frac{6}{\left(\frac{\sqrt{2}}{2}\right)^2}=12$$
$$y=\cos\left(\frac{\pi}{3}-2x\right), \quad x_0=\frac{\pi}{3}$$
$$y’=2\sin\left(\frac{\pi}{3}-2x\right)$$
$$y’\left(\frac{\pi}{3}\right)=2\sin\left(\frac{\pi}{3}-2\cdot\frac{\pi}{3}\right)=2\sin\left(-\frac{\pi}{3}\right)=-2\cdot\frac{\sqrt{3}}{2}=-\sqrt{3}$$
$$y=\ctg\frac{x}{3}, \quad x_0=\pi$$
$$y’=-\frac{1}{3\sin^2\frac{x}{3}}$$
$$y'(\pi)=-\frac{1}{3\sin^2\frac{\pi}{3}}=-\frac{1}{3\cdot\left(\frac{\sqrt{3}}{2}\right)^2}=-\frac{4}{9}$$
Ответ
а) $$0$$; б) $$12$$; в) $$-\sqrt{3}$$; г) $$-\frac{4}{9}$$.
