Вопрос:
218. Найдите производную тригонометрической функции:
Ответ:
Решение:
- \( y = \sin 3x \) \( y' = \cos 3x \cdot (3x)' = 3\cos 3x \)
- \( y = \cos 2x \) \( y' = -\sin 2x \cdot (2x)' = -2\sin 2x \)
- \( y = \operatorname{tg} x - x \) \( y' = \frac{1}{\cos^2 x} - 1 \)
- \( y = \operatorname{ctg} x + x \) \( y' = -\frac{1}{\sin^2 x} + 1 \)
- \( y = \operatorname{tg} x - \operatorname{ctg} x \) \( y' = \frac{1}{\cos^2 x} - \left(-\frac{1}{\sin^2 x}\right) = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x} = \frac{4}{(2\sin x \cos x)^2} = \frac{4}{\sin^2 2x} \)
- \( y = \sin t \cos t \) \( y = \frac{1}{2} \sin 2t \) \( y' = \frac{1}{2} \cos 2t \cdot 2 = \cos 2t \)
- \( y = \sin x(1 + \cos x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x \) \( y' = \cos x + \frac{1}{2} \cos 2x \cdot 2 = \cos x + \cos 2x \)
- \( v = \frac{1 - \cos t}{1 + \cos t} \) \( v = \frac{2\sin^2 \frac{t}{2}}{2\cos^2 \frac{t}{2}} = \operatorname{tg}^2 \frac{t}{2} \) \( v' = 2 \operatorname{tg} \frac{t}{2} \cdot \frac{1}{\cos^2 \frac{t}{2}} \cdot \frac{1}{2} = \frac{\operatorname{tg} \frac{t}{2}}{\cos^2 \frac{t}{2}} = \frac{\sin \frac{t}{2}}{\cos^3 \frac{t}{2}} \)
Похожие