$$$\frac{x^{3} - 2 x^{2}}{x^{2} + 1}$$$の導関数

$$$f{\left(x \right)} = x^{3} - 2 x^{2}$$$ 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：

$${\color{red}\left(\frac{d}{dx} \left(\frac{x^{3} - 2 x^{2}}{x^{2} + 1}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dx} \left(x^{3} - 2 x^{2}\right) \left(x^{2} + 1\right) - \left(x^{3} - 2 x^{2}\right) \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}\right)}$$

和/差の導関数は、導関数の和/差である:

$$\frac{\left(x^{2} + 1\right) {\color{red}\left(\frac{d}{dx} \left(x^{3} - 2 x^{2}\right)\right)} - \left(x^{3} - 2 x^{2}\right) \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}} = \frac{\left(x^{2} + 1\right) {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) - \frac{d}{dx} \left(2 x^{2}\right)\right)} - \left(x^{3} - 2 x^{2}\right) \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}$$

冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = 3$$$ に対して適用する:

$$\frac{\left(x^{2} + 1\right) \left({\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)} - \frac{d}{dx} \left(2 x^{2}\right)\right) - \left(x^{3} - 2 x^{2}\right) \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}} = \frac{\left(x^{2} + 1\right) \left({\color{red}\left(3 x^{2}\right)} - \frac{d}{dx} \left(2 x^{2}\right)\right) - \left(x^{3} - 2 x^{2}\right) \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}$$

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 $$$f{\left(x \right)} = x^{2}$$$ に対して適用します:

$$\frac{\left(x^{2} + 1\right) \left(3 x^{2} - {\color{red}\left(\frac{d}{dx} \left(2 x^{2}\right)\right)}\right) - \left(x^{3} - 2 x^{2}\right) \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}} = \frac{\left(x^{2} + 1\right) \left(3 x^{2} - {\color{red}\left(2 \frac{d}{dx} \left(x^{2}\right)\right)}\right) - \left(x^{3} - 2 x^{2}\right) \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}$$

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:

$$\frac{\left(x^{2} + 1\right) \left(3 x^{2} - 2 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)}\right) - \left(x^{3} - 2 x^{2}\right) \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}} = \frac{\left(x^{2} + 1\right) \left(3 x^{2} - 2 {\color{red}\left(2 x\right)}\right) - \left(x^{3} - 2 x^{2}\right) \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}$$

和/差の導関数は、導関数の和/差である:

$$\frac{\left(x^{2} + 1\right) \left(3 x^{2} - 4 x\right) - \left(x^{3} - 2 x^{2}\right) {\color{red}\left(\frac{d}{dx} \left(x^{2} + 1\right)\right)}}{\left(x^{2} + 1\right)^{2}} = \frac{\left(x^{2} + 1\right) \left(3 x^{2} - 4 x\right) - \left(x^{3} - 2 x^{2}\right) {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(1\right)\right)}}{\left(x^{2} + 1\right)^{2}}$$

定数の導数は$$$0$$$です:

$$\frac{\left(x^{2} + 1\right) \left(3 x^{2} - 4 x\right) - \left(x^{3} - 2 x^{2}\right) \left({\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x^{2}\right)\right)}{\left(x^{2} + 1\right)^{2}} = \frac{\left(x^{2} + 1\right) \left(3 x^{2} - 4 x\right) - \left(x^{3} - 2 x^{2}\right) \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)\right)}{\left(x^{2} + 1\right)^{2}}$$

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:

$$\frac{\left(x^{2} + 1\right) \left(3 x^{2} - 4 x\right) - \left(x^{3} - 2 x^{2}\right) {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)}}{\left(x^{2} + 1\right)^{2}} = \frac{\left(x^{2} + 1\right) \left(3 x^{2} - 4 x\right) - \left(x^{3} - 2 x^{2}\right) {\color{red}\left(2 x\right)}}{\left(x^{2} + 1\right)^{2}}$$

簡単化せよ:

$$\frac{- 2 x \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) \left(3 x^{2} - 4 x\right)}{\left(x^{2} + 1\right)^{2}} = \frac{x \left(x - 1\right) \left(x^{2} + x + 4\right)}{\left(x^{2} + 1\right)^{2}}$$

したがって、$$$\frac{d}{dx} \left(\frac{x^{3} - 2 x^{2}}{x^{2} + 1}\right) = \frac{x \left(x - 1\right) \left(x^{2} + x + 4\right)}{\left(x^{2} + 1\right)^{2}}$$$。