ГДЗ по Алгебре 10 Класс Номер 42.14 Профильный Уровень Мордкович — Подробные Ответы

а) $y = \sin\left(3x — \frac{\pi}{4}\right)$ и $x_0 = \frac{\pi}{4}$;

б) $y = \operatorname{tg} 6x$ и $x_0 = \frac{\pi}{24}$;

в) $y = \cos\left(\frac{\pi}{3} — 2x\right)$ и $x_0 = \frac{\pi}{3}$;

г) $y = \operatorname{ctg} \frac{x}{3}$ и $x_0 = \pi$

а) $y = \sin\left(3x — \frac{\pi}{4}\right)$ и $x_0 = \frac{\pi}{4}$;

Пусть $u = 3x — \frac{\pi}{4}$, тогда $y = \sin u$;

$y’ = (\sin u)’ \cdot \left(3x — \frac{\pi}{4}\right)’ = \cos u \cdot 3 = 3 \cos\left(3x — \frac{\pi}{4}\right)$;

$y'(x_0) = 3 \cos\left(\frac{3\pi}{4} — \frac{\pi}{4}\right) = 3 \cos \frac{\pi}{2} = 3 \cdot 0 = 0$;

б) $y = \operatorname{tg} 6x$ и $x_0 = \frac{\pi}{24}$;

Пусть $u = 6x$, тогда $y = \operatorname{tg} u$;

$y’ = (\operatorname{tg} u)’ \cdot (6x)’ = \frac{1}{\cos^2 u} \cdot 6 = \frac{6}{\cos^2 6x}$;

$y'(x_0) = \frac{6}{\cos^2 \frac{6\pi}{24}} = \frac{6}{\cos^2 \frac{\pi}{4}} = 6 : \left(\frac{\sqrt{2}}{2}\right)^2 = 6 \cdot \left(\frac{2}{\sqrt{2}}\right)^2 = 6 \cdot \frac{4}{2} = 12$;

в) $y = \cos\left(\frac{\pi}{3} — 2x\right)$ и $x_0 = \frac{\pi}{3}$;

Пусть $u = \frac{\pi}{3} — 2x$, тогда $y = \cos u$;

$y’ = (\cos u)’ \cdot \left(\frac{\pi}{3} — 2x\right)’ = -\sin u \cdot (-2) = 2 \sin\left(\frac{\pi}{3} — 2x\right)$;

$y'(x_0) = 2 \sin\left(\frac{\pi}{3} — \frac{2\pi}{3}\right) = 2 \sin\left(-\frac{\pi}{3}\right) = 2 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3}$;

г) $y = \operatorname{ctg} \frac{x}{3}$ и $x_0 = \pi$;

Пусть $u = \frac{x}{3}$, тогда $y = \operatorname{ctg} u$;

$y’ = (\operatorname{ctg} u)’ \cdot \left(\frac{x}{3}\right)’ = -\frac{1}{\sin^2 u} \cdot \frac{1}{3} = -\frac{1}{3 \sin^2 \left(\frac{x}{3}\right)}$;

$y'(x_0) = -\frac{1}{3 \sin^2 \frac{\pi}{3}} = -\frac{1}{3} : \left(\frac{\sqrt{3}}{2}\right)^2 = -\frac{1}{3} \cdot \left(\frac{2}{\sqrt{3}}\right)^2 = -\frac{1}{3} \cdot \frac{4}{3} = -\frac{4}{9}$

а) $y = \sin\left(3x — \dfrac{\pi}{4}\right)$, $x_0 = \dfrac{\pi}{4}$

Шаг 1. Вводим замену:

$$u = 3x — \frac{\pi}{4}, \quad \text{тогда } y = \sin u$$

Шаг 2. Используем цепное правило:

$$y’ = \frac{dy}{du} \cdot \frac{du}{dx}$$

Шаг 3. Производные:

• $\frac{dy}{du} = \cos u$
• $\frac{du}{dx} = 3$

Шаг 4. Собираем:

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

Шаг 5. Подставим $x = \frac{\pi}{4}$:

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

Ответ:

$$y’\left(\frac{\pi}{4}\right) = 0$$

б) $y = \operatorname{tg}(6x)$, $x_0 = \dfrac{\pi}{24}$

Шаг 1. Пусть:

$$u = 6x \Rightarrow y = \operatorname{tg}(u)$$

Шаг 2. Производная тангенса:

$$\frac{d}{du}(\operatorname{tg} u) = \frac{1}{\cos^2 u}$$

Шаг 3. Цепное правило:

$$y’ = \frac{1}{\cos^2 u} \cdot \frac{du}{dx} = \frac{1}{\cos^2(6x)} \cdot 6 = \frac{6}{\cos^2(6x)}$$

Шаг 4. Подставим $x = \frac{\pi}{24}$:

$$u = 6x = 6 \cdot \frac{\pi}{24} = \frac{\pi}{4}$$ $$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \Rightarrow \cos^2 \frac{\pi}{4} = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}$$ $$y’\left(\frac{\pi}{24}\right) = \frac{6}{\frac{1}{2}} = 12$$

Ответ:

$$y’\left(\frac{\pi}{24}\right) = 12$$

в) $y = \cos\left(\dfrac{\pi}{3} — 2x\right)$, $x_0 = \dfrac{\pi}{3}$

Шаг 1. Обозначим:

$$u = \frac{\pi}{3} — 2x, \quad y = \cos(u)$$

Шаг 2. Производная:

$$y’ = \frac{dy}{du} \cdot \frac{du}{dx} = -\sin(u) \cdot (-2) = 2 \sin\left(\frac{\pi}{3} — 2x\right)$$

Шаг 3. Подставим $x = \frac{\pi}{3}$:

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

Ответ:

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

г) $y = \operatorname{ctg}\left(\dfrac{x}{3}\right)$, $x_0 = \pi$

Шаг 1. Вводим замену:

$$u = \frac{x}{3}, \quad y = \operatorname{ctg}(u)$$

Шаг 2. Производная котангенса:

$$\frac{d}{du}(\operatorname{ctg} u) = -\frac{1}{\sin^2 u}$$

Шаг 3. Цепное правило:

$$y’ = -\frac{1}{\sin^2 u} \cdot \frac{du}{dx} = -\frac{1}{\sin^2 \left( \frac{x}{3} \right)} \cdot \frac{1}{3} = -\frac{1}{3 \sin^2 \left( \frac{x}{3} \right)}$$

Шаг 4. Подставим $x = \pi$:

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

Ответ:

$$y'(\pi) = -\frac{4}{9}$$