220. Вычислите производную тригонометрической функции в данном значении аргумента х:

Решение:

1. \( f(x) = \sin^2 2x \) \( f'(x) = 2\sin 2x \cdot \cos 2x \cdot 2 = 2\sin 4x \). При \( x = \frac{\pi}{16} \) \( f'(\frac{\pi}{16}) = 2\sin (4 \cdot \frac{\pi}{16}) = 2\sin \frac{\pi}{4} = 2\frac{\sqrt{2}}{2} = \sqrt{2} \)
2. \( f(x) = \cos^2 2x \) \( f'(x) = 2\cos 2x \cdot (-\sin 2x) \cdot 2 = -2\sin 4x \). При \( x = \frac{\pi}{16} \) \( f'(\frac{\pi}{16}) = -2\sin (4 \cdot \frac{\pi}{16}) = -2\sin \frac{\pi}{4} = -2\frac{\sqrt{2}}{2} = -\sqrt{2} \)
3. \( f(x) = \operatorname{tg}^2 3x \) \( f'(x) = 2\operatorname{tg} 3x \cdot \frac{1}{\cos^2 3x} \cdot 3 = \frac{6\operatorname{tg} 3x}{\cos^2 3x} \). При \( x = \frac{\pi}{12} \) \( f'(\frac{\pi}{12}) = \frac{6\operatorname{tg} (3 \cdot \frac{\pi}{12})}{\cos^2 (3 \cdot \frac{\pi}{12})} = \frac{6\operatorname{tg} \frac{\pi}{4}}{\cos^2 \frac{\pi}{4}} = \frac{6 \cdot 1}{(\frac{\sqrt{2}}{2})^2} = \frac{6}{1/2} = 12 \)
4. \( f(x) = \operatorname{tg}^2 x \sin x \) \( f'(x) = 2\operatorname{tg} x \cdot \frac{1}{\cos^2 x} \cdot \sin x + \operatorname{tg}^2 x \cdot \cos x = \frac{2\sin x}{\cos^3 x} + \frac{\sin^2 x \cos x}{\cos^2 x} = \frac{2\sin x + \sin^2 x \cos^2 x}{\cos^3 x} \). При \( x = \frac{\pi}{3} \) \( f'(\frac{\pi}{3}) = \frac{2\frac{\sqrt{3}}{2} + (\frac{\sqrt{3}}{2})^2 (\frac{1}{2})^2}{(\frac{1}{2})^3} = \frac{\sqrt{3} + \frac{3}{4} \cdot \frac{1}{4}}{\frac{1}{8}} = 8(\sqrt{3} + \frac{3}{16}) = 8\sqrt{3} + \frac{3}{2} \)
5. \( f(x) = 8 \sin^2 x \cos x \) \( f'(x) = 8 \cdot 2\sin x \cos x \cdot \cos x + 8\sin^2 x \cdot (-\sin x) = 16\sin x \cos^2 x - 8\sin^3 x \). При \( x = \frac{\pi}{4} \) \( f'(\frac{\pi}{4}) = 16\frac{\sqrt{2}}{2} (\frac{\sqrt{2}}{2})^2 - 8(\frac{\sqrt{2}}{2})^3 = 16\frac{\sqrt{2}}{2} \frac{1}{2} - 8 \frac{2\sqrt{2}}{8} = 4\sqrt{2} - \sqrt{2} = 3\sqrt{2} \)
6. \( f(x) = \operatorname{tg}^2 x - \operatorname{ctg}^2 x \) \( f'(x) = 2\operatorname{tg} x \cdot \frac{1}{\cos^2 x} - 2\operatorname{ctg} x \cdot \left(-\frac{1}{\sin^2 x}\right) = \frac{2\sin x}{\cos^3 x} + \frac{2\cos x}{\sin^3 x} \). При \( x = \frac{\pi}{4} \) \( f'(\frac{\pi}{4}) = \frac{2\frac{\sqrt{2}}{2}}{(\frac{\sqrt{2}}{2})^3} + \frac{2\frac{\sqrt{2}}{2}}{(\frac{\sqrt{2}}{2})^3} = \frac{\sqrt{2}}{1/2\sqrt{2}} + \frac{\sqrt{2}}{1/2\sqrt{2}} = 4 + 4 = 8 \)
7. \( f(x) = \cos 2x(1 + \sin 2x) = \cos 2x + \cos 2x \sin 2x = \cos 2x + \frac{1}{2}\sin 4x \) \( f'(x) = -2\sin 2x + \frac{1}{2}\cos 4x \cdot 4 = -2\sin 2x + 2\cos 4x \). При \( x = \frac{\pi}{8} \) \( f'(\frac{\pi}{8}) = -2\sin (2 \cdot \frac{\pi}{8}) + 2\cos (4 \cdot \frac{\pi}{8}) = -2\sin \frac{\pi}{4} + 2\cos \frac{\pi}{2} = -2\frac{\sqrt{2}}{2} + 2 \cdot 0 = -\sqrt{2} \)
8. \( f(x) = \frac{1 - \operatorname{tg} 2x}{\operatorname{tg} 2x} = \frac{1}{\operatorname{tg} 2x} - 1 = \operatorname{ctg} 2x - 1 \) \( f'(x) = -\frac{1}{\sin^2 2x} \cdot 2 = -\frac{2}{\sin^2 2x} \). При \( x = \frac{\pi}{4} \) \( f'(\frac{\pi}{4}) = -\frac{2}{\sin^2 (2 \cdot \frac{\pi}{4})} = -\frac{2}{\sin^2 \frac{\pi}{2}} = -\frac{2}{1^2} = -2 \)