$$$\cot{\left(\frac{x}{2} \right)}$$$的导数

函数$$$\cot{\left(\frac{x}{2} \right)}$$$是两个函数$$$f{\left(u \right)} = \cot{\left(u \right)}$$$和$$$g{\left(x \right)} = \frac{x}{2}$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

$${\color{red}\left(\frac{d}{dx} \left(\cot{\left(\frac{x}{2} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\cot{\left(u \right)}\right) \frac{d}{dx} \left(\frac{x}{2}\right)\right)}$$

余切函数的导数为$$$\frac{d}{du} \left(\cot{\left(u \right)}\right) = - \csc^{2}{\left(u \right)}$$$：

$${\color{red}\left(\frac{d}{du} \left(\cot{\left(u \right)}\right)\right)} \frac{d}{dx} \left(\frac{x}{2}\right) = {\color{red}\left(- \csc^{2}{\left(u \right)}\right)} \frac{d}{dx} \left(\frac{x}{2}\right)$$

返回到原变量：

$$- \csc^{2}{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(\frac{x}{2}\right) = - \csc^{2}{\left({\color{red}\left(\frac{x}{2}\right)} \right)} \frac{d}{dx} \left(\frac{x}{2}\right)$$

对 $$$c = \frac{1}{2}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

$$- \csc^{2}{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{d}{dx} \left(\frac{x}{2}\right)\right)} = - \csc^{2}{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{\frac{d}{dx} \left(x\right)}{2}\right)}$$

应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：

$$- \frac{\csc^{2}{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{2} = - \frac{\csc^{2}{\left(\frac{x}{2} \right)} {\color{red}\left(1\right)}}{2}$$

化简：

$$- \frac{\csc^{2}{\left(\frac{x}{2} \right)}}{2} = \frac{1}{\cos{\left(x \right)} - 1}$$

因此，$$$\frac{d}{dx} \left(\cot{\left(\frac{x}{2} \right)}\right) = \frac{1}{\cos{\left(x \right)} - 1}$$$。