$$$\frac{1}{x^{2} + 1}$$$の導関数

関数$$$\frac{1}{x^{2} + 1}$$$は、2つの関数$$$f{\left(u \right)} = \frac{1}{u}$$$と$$$g{\left(x \right)} = x^{2} + 1$$$の合成$$$f{\left(g{\left(x \right)} \right)}$$$である。

連鎖律 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ を適用する:

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

冪法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ を $$$n = -1$$$ に対して適用する:

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

元の変数に戻す:

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

和/差の導関数は、導関数の和/差である:

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

定数の導数は$$$0$$$です:

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

冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = 2$$$ に対して適用する:

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

したがって、$$$\frac{d}{dx} \left(\frac{1}{x^{2} + 1}\right) = - \frac{2 x}{\left(x^{2} + 1\right)^{2}}$$$。