$$$x \left(2 - 3 x\right)$$$ 的導數

將乘積法則 $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ 應用於 $$$f{\left(x \right)} = x$$$ 和 $$$g{\left(x \right)} = 2 - 3 x$$$：

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

和/差的導數等於導數的和/差：

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

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

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

常數的導數為$$$0$$$：

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

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 $$$f{\left(x \right)} = x$$$：

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

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

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

因此，$$$\frac{d}{dx} \left(x \left(2 - 3 x\right)\right) = 2 - 6 x$$$。