$$$\cos{\left(\frac{x}{2} \right)}$$$的导数
您的输入
求$$$\frac{d}{dx} \left(\cos{\left(\frac{x}{2} \right)}\right)$$$。
解答
函数$$$\cos{\left(\frac{x}{2} \right)}$$$是两个函数$$$f{\left(u \right)} = \cos{\left(u \right)}$$$和$$$g{\left(x \right)} = \frac{x}{2}$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(\cos{\left(\frac{x}{2} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{dx} \left(\frac{x}{2}\right)\right)}$$余弦函数的导数是$$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$：
$${\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{dx} \left(\frac{x}{2}\right) = {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{dx} \left(\frac{x}{2}\right)$$返回到原变量：
$$- \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(\frac{x}{2}\right) = - \sin{\left({\color{red}\left(\frac{x}{2}\right)} \right)} \frac{d}{dx} \left(\frac{x}{2}\right)$$对 $$$c = \frac{1}{2}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$$- \sin{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{d}{dx} \left(\frac{x}{2}\right)\right)} = - \sin{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{\frac{d}{dx} \left(x\right)}{2}\right)}$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：
$$- \frac{\sin{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{2} = - \frac{\sin{\left(\frac{x}{2} \right)} {\color{red}\left(1\right)}}{2}$$因此，$$$\frac{d}{dx} \left(\cos{\left(\frac{x}{2} \right)}\right) = - \frac{\sin{\left(\frac{x}{2} \right)}}{2}$$$。
答案
$$$\frac{d}{dx} \left(\cos{\left(\frac{x}{2} \right)}\right) = - \frac{\sin{\left(\frac{x}{2} \right)}}{2}$$$A