$$$x^{2} \left(1 - x\right)$$$ 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 $$$f{\left(x \right)} = x^{2}$$$ 和 $$$g{\left(x \right)} = 1 - x$$$：

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

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

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

和/差的導數等於導數的和/差：

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

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

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

常數的導數為$$$0$$$：

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

化簡：

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

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