第6章 6.12

(1)

$y' = \cos x + x(-\sin x) = \cos x - x \sin x$

(2)

$y' = 2x \cdot \sin \frac{1}{x} + x^2 \cdot \cos \frac{1}{x} \cdot \left(-\frac{1}{x^2}\right) = 2x \cdot \sin \frac{1}{x} - \cos \frac{1}{x}$

(3)

$y' = \frac{-\cos x \cdot (1+\sin x) - (1-\sin x) \cdot \cos x}{(1+\sin x)^2} = -\frac{2\cos x}{(1+\sin x)^2}$

(4)

$y' = \frac{2\cos 2x \cdot (1+\cos 2x) - \sin 2x \cdot (-2\sin 2x)}{(1+\cos 2x)^2} = \frac{2\cos 2x + 2\cos^2 2x + 2\sin^2 2x}{(1+\cos 2x)^2} = \frac{2\cos 2x + 2}{(1+\cos 2x)^2} = \frac{2}{1+\cos 2x}$

(5)

$y' = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2} = \frac{-2}{(\sin x - \cos x)^2}$

(6)

$y = \log\left(\tan \frac{x}{2}\right) = \log\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right) = \log\left(\sin \frac{x}{2}\right) - \log\left(\cos \frac{x}{2}\right)$

$y' = \frac{\frac{1}{2}\cos \frac{x}{2}}{\sin \frac{x}{2}} + \frac{\frac{1}{2}\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{1}{2} \cdot \frac{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}}{\sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{2\sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{\sin x}$