$$$\frac{x^{3} - 2 x^{2}}{x^{2} + 1}$$$的导数

对 $$$f{\left(x \right)} = x^{3} - 2 x^{2}$$$ 和 $$$g{\left(x \right)} = x^{2} + 1$$$ 应用商法则 $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$：

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

和/差的导数等于导数的和/差：

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

应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = 3$$$:

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

对 $$$c = 2$$$ 和 $$$f{\left(x \right)} = x^{2}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

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

应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = 2$$$:

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

和/差的导数等于导数的和/差：

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

常数的导数是$$$0$$$:

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

应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = 2$$$:

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

化简：

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

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