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.

1. $$y=\sin\left(3x-\frac{\pi}{4}\right), \quad x_0=\frac{\pi}{4}$$

   $$y’ = 3\cos\left(3x-\frac{\pi}{4}\right)$$

   $$y’\left(\frac{\pi}{4}\right)=3\cos\left(3\cdot\frac{\pi}{4}-\frac{\pi}{4}\right)=3\cos\frac{\pi}{2}=0$$

2. $$y=\tg 6x, \quad x_0=\frac{\pi}{24}$$

   $$y’=\frac{6}{\cos^2 6x}$$

   $$y’\left(\frac{\pi}{24}\right)=\frac{6}{\cos^2\left(6\cdot\frac{\pi}{24}\right)}=\frac{6}{\cos^2\frac{\pi}{4}}=\frac{6}{\left(\frac{\sqrt{2}}{2}\right)^2}=12$$

3. $$y=\cos\left(\frac{\pi}{3}-2x\right), \quad x_0=\frac{\pi}{3}$$

   $$y’=2\sin\left(\frac{\pi}{3}-2x\right)$$

   $$y’\left(\frac{\pi}{3}\right)=2\sin\left(\frac{\pi}{3}-2\cdot\frac{\pi}{3}\right)=2\sin\left(-\frac{\pi}{3}\right)=-2\cdot\frac{\sqrt{3}}{2}=-\sqrt{3}$$

4. $$y=\ctg\frac{x}{3}, \quad x_0=\pi$$

   $$y’=-\frac{1}{3\sin^2\frac{x}{3}}$$

   $$y'(\pi)=-\frac{1}{3\sin^2\frac{\pi}{3}}=-\frac{1}{3\cdot\left(\frac{\sqrt{3}}{2}\right)^2}=-\frac{4}{9}$$

Ответ

а) $$0$$; б) $$12$$; в) $$-\sqrt{3}$$; г) $$-\frac{4}{9}$$.