ГДЗ по Алгебре 10 Класс Номер 42.16 Профильный Уровень Мордкович — Подробные Ответы

а) $y = (x — \sin x)^2$ и $x_0 = \pi$;

б) $y = \sqrt{\frac{1 — \sin x}{\cos x}}$, и $x_0 = \frac{\pi}{4}$;

в) $y = \sqrt{(\sin x + 1)\cos x}$, и $x_0 = \frac{\pi}{6}$;

г) $y = (\tg x — 1)^4$, и $x_0 = \frac{\pi}{4}$

а) $y = (x — \sin x)^2$ и $x_0 = \pi$;

Пусть $u = x — \sin x$, тогда $y = u^2$;

$$y’ = (u^2)’ \cdot (x — \sin x)’ = 2u \cdot (1 — \cos x) = 2(x — \sin x)(1 — \cos x);$$ $$y'(x_0) = 2(\pi — \sin \pi)(1 — \cos \pi) = 2(\pi — 0)(1 — (-1)) = 2\pi \cdot 2 = 4\pi;$$

б) $y = \sqrt{\frac{1 — \sin x}{\cos x}}$, и $x_0 = \frac{\pi}{4}$;

Пусть $u = \frac{1 — \sin x}{\cos x}$, тогда $y = \sqrt{u}$;

$$u’_x = \frac{(1 — \sin x)’ \cos x — (1 — \sin x)(\cos x)’}{\cos^2 x} =$$ $$= \frac{(0 — \cos x)\cos x — (1 — \sin x)(- \sin x)}{\cos^2 x} = \frac{-\cos^2 x + \sin x — \sin^2 x}{\cos^2 x} =$$ $$= \frac{1 + \sin x}{\cos x} — \tg^2 x;$$

$$y’ = (\sqrt{u})’ \cdot u’_x = \frac{u’_x}{2\sqrt{u}} = \frac{\frac{\tg x}{\cos x} — 1 — \tg^2 x}{2\sqrt{\frac{1 — \sin x}{\cos x}}};$$

$$y'(x_0) = \frac{\frac{\tg \frac{\pi}{4}}{\cos \frac{\pi}{4}} — 1 — \tg^2 \frac{\pi}{4}}{2 \sqrt{\frac{1 — \sin \frac{\pi}{4}}{\cos \frac{\pi}{4}}}} = \frac{1 : \frac{\sqrt{2}}{2} — 1 — 1^2}{2 \sqrt{\frac{1 — \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}}} =$$ $$= \frac{\frac{2}{\sqrt{2}} — 2}{2 \sqrt{\frac{2 — \sqrt{2}}{\sqrt{2}}}} = \frac{2\sqrt{2} — 2\sqrt{2}}{4\sqrt{2 — \sqrt{2}}} = -\frac{1}{2} \sqrt{2(\sqrt{2} — 1)};$$

в) $y = \sqrt{(\sin x + 1)\cos x}$, и $x_0 = \frac{\pi}{6}$;

Пусть $u = (\sin x + 1)\cos x$, тогда $y = \sqrt{u}$;

$$u’_x = (\sin x + 1)’ \cos x + (\sin x + 1)(\cos x)’ = (\cos x + 0) \cos x + (\sin x + 1)(- \sin x) =$$ $$= \cos^2 x — \sin^2 x — \sin x == \cos 2x — \sin x;$$

$$y’ = (\sqrt{u})’ \cdot u’_x = \frac{u’_x}{2\sqrt{u}} = \frac{\cos 2x — \sin x}{2 \sqrt{(\sin x + 1)\cos x}};$$

$$y'(x_0) = \frac{\cos \frac{\pi}{3} — \sin \frac{\pi}{6}}{2 \sqrt{(\sin \frac{\pi}{6} + 1) \cos \frac{\pi}{6}}} = \frac{\frac{1}{2} — \frac{1}{2}}{2 \sqrt{(\frac{1}{2} + 1) \cdot \frac{\sqrt{3}}{2}}} = 0;$$

г) $y = (\tg x — 1)^4$, и $x_0 = \frac{\pi}{4}$;

Пусть $u = \tg x — 1$, тогда $y = u^4$;

$$y’ = (u^4)’ = 4u^3 \cdot (\tg x — 1)’ = 4u^3 \cdot \frac{1}{\cos^2 x} = \frac{4(\tg x — 1)^3}{\cos^2 x};$$ $$y'(x_0) = \frac{4 \cdot (\tg \frac{\pi}{4} — 1)^3}{\cos^2 \frac{\pi}{4}} = \frac{4 \cdot (1 — 1)^3}{(\sqrt{2}/2)^2} = \frac{4 \cdot 0}{2/4} = 0;$$

а) $y = (x — \sin x)^2$, $x_0 = \pi$

1. Заменим выражение под квадратом:

Пусть $u = x — \sin x$, тогда

$$y = u^2$$

2. Продифференцируем y:

$$y’ = \frac{d}{dx}(u^2) = 2u \cdot u’$$

3. Найдём производную u:

$$u = x — \sin x \Rightarrow u’ = \frac{d}{dx}(x — \sin x) = 1 — \cos x$$

4. Подставим u и u’ в y’:

$$y’ = 2(x — \sin x)(1 — \cos x)$$

5. Подставим $x_0 = \pi$:

$$\sin \pi = 0, \quad \cos \pi = -1$$ $$y'(\pi) = 2(\pi — 0)(1 — (-1)) = 2\pi \cdot 2 = \boxed{4\pi}$$

б) $y = \sqrt{\dfrac{1 — \sin x}{\cos x}}, \quad x_0 = \dfrac{\pi}{4}$

1. Обозначим подкоренное выражение как u:

$$u = \frac{1 — \sin x}{\cos x} \Rightarrow y = \sqrt{u}$$

2. Найдём производную u (по правилу дроби):

$$u = \frac{f(x)}{g(x)}, \quad f(x) = 1 — \sin x, \quad g(x) = \cos x$$ $$u’ = \frac{f'(x) \cdot g(x) — f(x) \cdot g'(x)}{g(x)^2}$$ $$f'(x) = -\cos x, \quad g'(x) = -\sin x$$ $$u’ = \frac{(-\cos x)\cos x — (1 — \sin x)(- \sin x)}{\cos^2 x}$$ $$= \frac{-\cos^2 x + \sin x — \sin^2 x}{\cos^2 x}$$

3. Упростим числитель:

$$-\cos^2 x + \sin x — \sin^2 x = \sin x — (\cos^2 x + \sin^2 x) = \sin x — 1$$ $$u’ = \frac{\sin x — 1}{\cos^2 x}$$

4. Теперь найдём производную y:

$$y = \sqrt{u} \Rightarrow y’ = \frac{1}{2\sqrt{u}} \cdot u’$$ $$y’ = \frac{\sin x — 1}{2\cos^2 x \cdot \sqrt{\frac{1 — \sin x}{\cos x}}}$$

5. Подставим $x = \frac{\pi}{4}$:

$$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

Подкоренное выражение:

$$u = \frac{1 — \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 — \sqrt{2}}{\sqrt{2}}$$

u’:

$$\sin x — 1 = \frac{\sqrt{2}}{2} — 1, \quad \cos^2 x = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}$$

6. Подставим в y’:

$$y’ = \frac{\frac{\sqrt{2}}{2} — 1}{2 \cdot \frac{1}{2} \cdot \sqrt{\frac{2 — \sqrt{2}}{\sqrt{2}}}} = \frac{\frac{\sqrt{2}}{2} — 1}{\sqrt{\frac{2 — \sqrt{2}}{\sqrt{2}}}} =$$ $$= \frac{(\sqrt{2} — 2)}{2 \cdot \sqrt{\frac{2 — \sqrt{2}}{\sqrt{2}}}} = -\frac{1}{2} \sqrt{2(\sqrt{2} — 1)} = \boxed{-\frac{1}{2} \sqrt{2(\sqrt{2} — 1)}}$$

в) $y = \sqrt{(\sin x + 1)\cos x}, \quad x_0 = \frac{\pi}{6}$

1. Обозначим подкоренное выражение как u:

$$u = (\sin x + 1)\cos x \Rightarrow y = \sqrt{u}$$

2. Найдём u’:

$$u = f(x) \cdot g(x), \quad f(x) = \sin x + 1, \quad g(x) = \cos x$$ $$f'(x) = \cos x, \quad g'(x) = -\sin x$$ $$u’ = f'(x) \cdot g(x) + f(x) \cdot g'(x) = \cos x \cdot \cos x + (\sin x + 1)(- \sin x)$$ $$= \cos^2 x — \sin^2 x — \sin x = \cos 2x — \sin x$$

3. Производная y:

$$y’ = \frac{1}{2\sqrt{u}} \cdot u’ = \frac{\cos 2x — \sin x}{2 \sqrt{(\sin x + 1)\cos x}}$$

4. Подставим $x = \frac{\pi}{6}$:

$$\sin \frac{\pi}{6} = \frac{1}{2}, \quad \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \cos \frac{\pi}{3} = \frac{1}{2}$$ $$\cos 2x = \cos \frac{\pi}{3} = \frac{1}{2}, \quad \sin \frac{\pi}{6} = \frac{1}{2}$$ $$\Rightarrow \text{Числитель: } \frac{1}{2} — \frac{1}{2} = 0$$ $$\Rightarrow y'(x_0) = 0$$ $$\boxed{y'(x_0) = 0}$$

г) $y = (\tg x — 1)^4, \quad x_0 = \frac{\pi}{4}$

1. Обозначим:

$$u = \tg x — 1 \Rightarrow y = u^4$$

2. Найдём u’:

$$u’ = \frac{d}{dx}(\tg x — 1) = \frac{1}{\cos^2 x}$$

3. Найдём y’:

$$y’ = 4u^3 \cdot u’ = \frac{4(\tg x — 1)^3}{\cos^2 x}$$

4. Подставим $x = \frac{\pi}{4}$:

$$\tg \frac{\pi}{4} = 1 \Rightarrow (\tg x — 1)^3 = 0^3 = 0$$ $$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \Rightarrow \cos^2 \frac{\pi}{4} = \frac{1}{2}$$ $$y’ = \frac{4 \cdot 0}{1/2} = 0$$ $$\boxed{y'(x_0) = 0}$$