2.求下列函数在指定点处的导数:-|||-(1) (x)=xsin x+dfrac (1)(2)cos x =dfrac (pi )(4)-|||-(2) (x)=dfrac (x-sin x)(x+sin x) =dfrac (pi )(2)

步骤 1：求导数
(1) 对于函数 $f(x)=x\sin x+\dfrac {1}{2}\cos x$，我们首先需要求出它的导数 $f'(x)$。
$f'(x) = \dfrac{d}{dx}(x\sin x) + \dfrac{d}{dx}(\dfrac{1}{2}\cos x)$
$= \sin x + x\cos x - \dfrac{1}{2}\sin x$
$= \dfrac{1}{2}\sin x + x\cos x$

步骤 2：代入指定点
将 $x=\dfrac{\pi}{4}$ 代入 $f'(x)$ 中，得到 $f'(\dfrac{\pi}{4})$。
$f'(\dfrac{\pi}{4}) = \dfrac{1}{2}\sin(\dfrac{\pi}{4}) + \dfrac{\pi}{4}\cos(\dfrac{\pi}{4})$
$= \dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2} + \dfrac{\pi}{4}\cdot\dfrac{\sqrt{2}}{2}$
$= \dfrac{\sqrt{2}}{4} + \dfrac{\pi\sqrt{2}}{8}$
$= \dfrac{2\sqrt{2} + \pi\sqrt{2}}{8}$
$= \dfrac{(\pi + 2)\sqrt{2}}{8}$

步骤 3：求导数
(2) 对于函数 $f(x)=\dfrac{x-\sin x}{x+\sin x}$，我们首先需要求出它的导数 $f'(x)$。
$f'(x) = \dfrac{(x+\sin x)\dfrac{d}{dx}(x-\sin x) - (x-\sin x)\dfrac{d}{dx}(x+\sin x)}{(x+\sin x)^2}$
$= \dfrac{(x+\sin x)(1-\cos x) - (x-\sin x)(1+\cos x)}{(x+\sin x)^2}$
$= \dfrac{x - x\cos x + \sin x - \sin x\cos x - x - x\cos x + \sin x + \sin x\cos x}{(x+\sin x)^2}$
$= \dfrac{-2x\cos x + 2\sin x}{(x+\sin x)^2}$

步骤 4：代入指定点
将 $x=\dfrac{\pi}{2}$ 代入 $f'(x)$ 中，得到 $f'(\dfrac{\pi}{2})$。
$f'(\dfrac{\pi}{2}) = \dfrac{-2\cdot\dfrac{\pi}{2}\cos(\dfrac{\pi}{2}) + 2\sin(\dfrac{\pi}{2})}{(\dfrac{\pi}{2}+\sin(\dfrac{\pi}{2}))^2}$
$= \dfrac{0 + 2}{(\dfrac{\pi}{2}+1)^2}$
$= \dfrac{2}{(\dfrac{\pi}{2}+1)^2}$
$= \dfrac{8}{(\pi+2)^2}$