$$$x^{3} - 3 x^{2}$$$ 的導數

和/差的導數等於導數的和/差：

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

套用常數倍法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$，使用 $$$c = 3$$$ 與 $$$f{\left(x \right)} = x^{2}$$$：

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

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

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

套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 3$$$：

$$- 6 x + {\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)} = - 6 x + {\color{red}\left(3 x^{2}\right)}$$

化簡：

$$3 x^{2} - 6 x = 3 x \left(x - 2\right)$$

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