219. Найдите производную тригонометрической функции:

Решение:

1. \( y = \sin^2 2x \) \( y = (\sin 2x)^2 \) \( y' = 2\sin 2x \cdot \cos 2x \cdot 2 = 2\sin 4x \)
2. \( y = \cos^2 x \) \( y = (\cos x)^2 \) \( y' = 2\cos x \cdot (-\sin x) = -\sin 2x \)
3. \( y = \operatorname{tg}^2 x \) \( y = (\operatorname{tg} x)^2 \) \( y' = 2\operatorname{tg} x \cdot \frac{1}{\cos^2 x} \)
4. \( y = \operatorname{ctg}^2 x \) \( y = (\operatorname{ctg} x)^2 \) \( y' = 2\operatorname{ctg} x \cdot \left(-\frac{1}{\sin^2 x}\right) = -\frac{2\cos x}{\sin^3 x} \)
5. \( y = \sin^2 2x \) \( y = (\sin 2x)^2 \) \( y' = 2\sin 2x \cdot \cos 2x \cdot 2 = 2\sin 4x \)
6. \( S = \sqrt{\sin 2\varphi} \) \( S' = \frac{1}{2\sqrt{\sin 2\varphi}} \cdot \cos 2\varphi \cdot 2 = \frac{\cos 2\varphi}{\sqrt{\sin 2\varphi}} \)
7. \( y = \sqrt[3]{\sin^2 2x} = (\sin^2 2x)^{1/3} \) \( y' = \frac{1}{3}(\sin^2 2x)^{-2/3} \cdot 2\sin 2x \cdot \cos 2x \cdot 2 = \frac{4\sin 2x \cos 2x}{3(\sin^2 2x)^{2/3}} = \frac{2\sin 4x}{3\sqrt[3]{\sin^4 2x}} \)
8. \( y = \frac{1}{5} \operatorname{tg}^5 x + \frac{2}{3} \operatorname{tg}^3 x \) \( y' = \frac{1}{5} \cdot 5\operatorname{tg}^4 x \cdot \frac{1}{\cos^2 x} + \frac{2}{3} \cdot 3\operatorname{tg}^2 x \cdot \frac{1}{\cos^2 x} = \frac{\operatorname{tg}^4 x}{\cos^2 x} + \frac{2\operatorname{tg}^2 x}{\cos^2 x} = \frac{\operatorname{tg}^2 x}{\cos^2 x}(\operatorname{tg}^2 x + 2) \)