Упр.12.2 ГДЗ Мордкович Семенов 11 класс (Алгебра)

а) y = arcsin(x), x = -v3/2; в) y = arccos(x), x = 1/2;
б) y = arctg(x), x = v3/3; г) y = arcctg(x), x = -v3. Найдите производную функции:
а) y=((x-3)/2)^2; в) y=((-x-1)/3)^3; д) y=((2x+1)/15)^3;
б) y=((4-x)/3)^2; г) y=((x+3)/6)^3; е) y=((2-5x)/4)^2.

1. Уравнение касательной.

   а) $$y=\arcsin x,\quad x_0=-\frac{\sqrt3}{2}$$

   $$y_0=\arcsin\left(-\frac{\sqrt3}{2}\right)=-\frac{\pi}{3}$$

   $$y’=\frac{1}{\sqrt{1-x^2}},\quad y'(x_0)=\frac{1}{\sqrt{1-\frac34}}=2$$

   $$y+\frac{\pi}{3}=2\left(x+\frac{\sqrt3}{2}\right)$$

   $$y=2x-\frac{\pi}{3}+\sqrt3$$

2. б) $$y=\arctg x,\quad x_0=\frac{\sqrt3}{3}$$

   $$y_0=\arctg\left(\frac{\sqrt3}{3}\right)=\frac{\pi}{6}$$

   $$y’=\frac{1}{1+x^2},\quad y'(x_0)=\frac{1}{1+\frac13}=\frac34$$

   $$y-\frac{\pi}{6}=\frac34\left(x-\frac{\sqrt3}{3}\right)$$

   $$y=\frac34x+\frac{\pi}{6}-\frac{\sqrt3}{4}$$

3. в) $$y=\arccos x,\quad x_0=\frac12$$

   $$y_0=\arccos\left(\frac12\right)=\frac{\pi}{3}$$

   $$y’=-\frac{1}{\sqrt{1-x^2}},\quad y'(x_0)=-\frac{1}{\sqrt{1-\frac14}}=-\frac{2\sqrt3}{3}$$

   $$y-\frac{\pi}{3}=-\frac{2\sqrt3}{3}\left(x-\frac12\right)$$

   $$y=-\frac{2\sqrt3}{3}x+\frac{\pi}{3}+\frac{\sqrt3}{3}$$

4. г) $$y=\arcctg x,\quad x_0=-\sqrt3$$

   $$y_0=\arcctg(-\sqrt3)=\pi-\arcctg\sqrt3=\frac{5\pi}{6}$$

   $$y’=-\frac{1}{1+x^2},\quad y'(x_0)=-\frac{1}{1+3}=-\frac14$$

   $$y-\frac{5\pi}{6}=-\frac14(x+\sqrt3)$$

   $$y=-\frac14x+\frac{5\pi}{6}-\frac{\sqrt3}{4}$$

5. Производная функции.

   а) $$y=\left(\frac{x-3}{2}\right)^2$$

   $$y’=\frac12\cdot 2\left(\frac{x-3}{2}\right)=\frac{x-3}{2}$$

6. б) $$y=\left(\frac{4-x}{3}\right)^2$$

   $$y’=-\frac13\cdot 2\left(\frac{4-x}{3}\right)=\frac{2(x-4)}{9}$$

7. в) $$y=\left(\frac{-x-1}{3}\right)^3$$

   $$y’=\frac13\cdot 3\left(\frac{-x-1}{3}\right)^2\cdot(-1)=-\left(\frac{x+1}{3}\right)^2$$

8. г) $$y=\left(\frac{x+3}{6}\right)^3$$

   $$y’=\frac16\cdot 3\left(\frac{x+3}{6}\right)^2=\frac12\left(\frac{x+3}{6}\right)^2$$

9. д) $$y=\left(\frac{2x+1}{15}\right)^3$$

   $$y’=\frac{2}{15}\cdot 3\left(\frac{2x+1}{15}\right)^2=\frac{2(2x+1)^2}{1125}$$

10. е) $$y=\left(\frac{2-5x}{4}\right)^2$$

    $$y’=-\frac54\cdot 2\left(\frac{2-5x}{4}\right)=\frac{5(5x-2)}{8}$$

Ответ

а) $$y=2x-\frac{\pi}{3}+\sqrt3$$;
б) $$y=\frac34x+\frac{\pi}{6}-\frac{\sqrt3}{4}$$;
в) $$y=-\frac{2\sqrt3}{3}x+\frac{\pi}{3}+\frac{\sqrt3}{3}$$;
г) $$y=-\frac14x+\frac{5\pi}{6}-\frac{\sqrt3}{4}$$.

а) $$y’=\frac{x-3}{2}$$;
б) $$y’=\frac{2(x-4)}{9}$$;
в) $$y’=-\left(\frac{x+1}{3}\right)^2$$;
г) $$y’=\frac12\left(\frac{x+3}{6}\right)^2$$;
д) $$y’=\frac{2(2x+1)^2}{1125}$$;
е) $$y’=\frac{5(5x-2)}{8}$$.