Turunan dari $$$x \left(x - 128\right)$$$

Terapkan aturan hasil kali $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ pada $$$f{\left(x \right)} = x$$$ dan $$$g{\left(x \right)} = x - 128$$$:

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

Turunan dari jumlah/selisih adalah jumlah/selisih dari turunan:

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

Turunan dari suatu konstanta adalah $$$0$$$:

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

Terapkan aturan pangkat $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ dengan $$$n = 1$$$, dengan kata lain, $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

Dengan demikian, $$$\frac{d}{dx} \left(x \left(x - 128\right)\right) = 2 x - 128$$$.