Упр.11.8 ГДЗ Мордкович Семенов 11 класс (Алгебра)

а) y=sin(x), x_0=?/3; г) y=cos(x), x_0=-?/2;
б) y=2cos(x), x_0=5?/6; д) y=-2sin(x), x_0=3?/4;
в) y=3tg(x), x_0=2?/3; е) y=-ctg(x), x_0=?/4.

1. $$y=\sin x$$

   $$y’=\cos x$$

   $$y’\left(\frac{\pi}{3}\right)=\cos\frac{\pi}{3}=\frac12$$

2. $$y=2\cos x$$

   $$y’=-2\sin x$$

   $$y’\left(\frac{5\pi}{6}\right)=-2\sin\frac{5\pi}{6}=-2\cdot\frac12=-1$$

3. $$y=3\tg x$$

   $$y’=\frac{3}{\cos^2 x}$$

   $$y’\left(\frac{2\pi}{3}\right)=\frac{3}{\cos^2\frac{2\pi}{3}}=\frac{3}{\left(-\frac12\right)^2}=12$$

4. $$y=\cos x$$

   $$y’=-\sin x$$

   $$y’\left(-\frac{\pi}{2}\right)=-\sin\left(-\frac{\pi}{2}\right)=1$$

5. $$y=-2\sin x$$

   $$y’=-2\cos x$$

   $$y’\left(\frac{3\pi}{4}\right)=-2\cos\frac{3\pi}{4}=-2\cdot\left(-\frac{\sqrt2}{2}\right)=\sqrt2$$

6. $$y=-\ctg x$$

   $$y’=\frac{1}{\sin^2 x}$$

   $$y’\left(\frac{\pi}{4}\right)=\frac{1}{\sin^2\frac{\pi}{4}}=\frac{1}{\left(\frac{\sqrt2}{2}\right)^2}=2$$

Ответ

а) $$\frac12$$; б) $$-1$$; в) $$12$$; г) $$1$$; д) $$\sqrt2$$; е) $$2$$.