$$$\frac{1}{x^{2} + 1}$$$ 的導數

函數 $$$\frac{1}{x^{2} + 1}$$$ 是兩個函數 $$$f{\left(u \right)} = \frac{1}{u}$$$ 與 $$$g{\left(x \right)} = x^{2} + 1$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。

應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

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

套用冪次法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，取 $$$n = -1$$$：

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

返回原變數：

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

和/差的導數等於導數的和/差：

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

常數的導數為$$$0$$$：

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x^{2}\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{{\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)}{\left(x^{2} + 1\right)^{2}}$$

套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 2$$$：

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

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