$$$- \frac{2 x}{x^{2} + 1}$$$の導関数

定数倍の法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ を $$$c = -2$$$ と $$$f{\left(x \right)} = \frac{x}{x^{2} + 1}$$$ に対して適用します:

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

$$$f{\left(x \right)} = x$$$ と $$$g{\left(x \right)} = x^{2} + 1$$$ に対して商の微分法則 $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$ を適用する：

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

$$$n = 1$$$ を用いて冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を適用すると、すなわち $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

和/差の導関数は、導関数の和/差である:

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

定数の導数は$$$0$$$です:

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

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:

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

簡単化せよ:

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

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