$$$\operatorname{atan}{\left(\frac{1}{x} \right)}$$$的导数

函数$$$\operatorname{atan}{\left(\frac{1}{x} \right)}$$$是两个函数$$$f{\left(u \right)} = \operatorname{atan}{\left(u \right)}$$$和$$$g{\left(x \right)} = \frac{1}{x}$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

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

$${\color{red}\left(\frac{d}{dx} \left(\operatorname{atan}{\left(\frac{1}{x} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right) \frac{d}{dx} \left(\frac{1}{x}\right)\right)}$$

反正切函数的导数为 $$$\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right) = \frac{1}{u^{2} + 1}$$$:

$${\color{red}\left(\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right)\right)} \frac{d}{dx} \left(\frac{1}{x}\right) = {\color{red}\left(\frac{1}{u^{2} + 1}\right)} \frac{d}{dx} \left(\frac{1}{x}\right)$$

返回到原变量：

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

应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = -1$$$:

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

化简：

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

因此，$$$\frac{d}{dx} \left(\operatorname{atan}{\left(\frac{1}{x} \right)}\right) = - \frac{1}{x^{2} + 1}$$$。