$$$x^{2} \left(1 - x\right)$$$の導関数

積の微分法 $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ を $$$f{\left(x \right)} = x^{2}$$$ と $$$g{\left(x \right)} = 1 - x$$$ に適用する:

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

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:

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

和/差の導関数は、導関数の和/差である:

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

$$$n = 1$$$ を用いて冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を適用すると、すなわち $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

定数の導数は$$$0$$$です:

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

簡単化せよ:

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

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