$$$x \left(x - 128\right)$$$的导数

对 $$$f{\left(x \right)} = x$$$ 和 $$$g{\left(x \right)} = x - 128$$$ 应用乘积法则 $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(x \left(x - 128\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) \left(x - 128\right) + x \frac{d}{dx} \left(x - 128\right)\right)}$$

和/差的导数等于导数的和/差：

$$x {\color{red}\left(\frac{d}{dx} \left(x - 128\right)\right)} + \left(x - 128\right) \frac{d}{dx} \left(x\right) = x {\color{red}\left(\frac{d}{dx} \left(x\right) - \frac{d}{dx} \left(128\right)\right)} + \left(x - 128\right) \frac{d}{dx} \left(x\right)$$

常数的导数是$$$0$$$:

$$x \left(- {\color{red}\left(\frac{d}{dx} \left(128\right)\right)} + \frac{d}{dx} \left(x\right)\right) + \left(x - 128\right) \frac{d}{dx} \left(x\right) = x \left(- {\color{red}\left(0\right)} + \frac{d}{dx} \left(x\right)\right) + \left(x - 128\right) \frac{d}{dx} \left(x\right)$$

应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：

$$x {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \left(x - 128\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = x {\color{red}\left(1\right)} + \left(x - 128\right) {\color{red}\left(1\right)}$$

因此，$$$\frac{d}{dx} \left(x \left(x - 128\right)\right) = 2 x - 128$$$。