$$$\left(x - 1\right)^{2}$$$的导数

函数$$$\left(x - 1\right)^{2}$$$是两个函数$$$f{\left(u \right)} = u^{2}$$$和$$$g{\left(x \right)} = x - 1$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

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

应用幂次法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，其中 $$$n = 2$$$:

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

返回到原变量：

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

和/差的导数等于导数的和/差：

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

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

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

常数的导数是$$$0$$$:

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

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