第6章 6.6

(1)

$y' = 3\cos 3x$

(2)

$y' = -4\sin 4x$

(3)

$y' = \dfrac{4}{\cos^2 4x}$

(4)

$y' = 2 \cdot 5 \cdot \sin(5x-7) = 10\sin(5x-7)$

(5)

$y = \tan^2 x = \dfrac{\sin^2 x}{\cos^2 x} = \dfrac{1-\cos^2 x}{\cos^2 x} = \dfrac{1}{\cos^2 x} - 1 = \dfrac{1}{\dfrac{1+\cos 2x}{2}} - 1 = \dfrac{2}{1+\cos 2x} - 1$

$y' = \dfrac{-2(-2\sin 2x)}{(1+\cos 2x)^2} = \dfrac{4\sin 2x}{(1+\cos 2x)^2} = \dfrac{\sin 2x}{\left(\dfrac{1+\cos 2x}{2}\right)^2} = \dfrac{\sin 2x}{\cos^4 x} = \dfrac{2\sin x \cos x}{\cos^4 x} = \dfrac{2\sin x}{\cos^3 x}$

(6)

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

(7)

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

(8)

$y' = \dfrac{-2\sin 2x}{2\sqrt{1+\cos 2x}} = -\dfrac{\sin 2x}{\sqrt{1+\cos 2x}}$