$$$\ln\left(1 - x\right)$$$的导数

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

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

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

自然对数的导数为 $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$：

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

返回到原变量：

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

和/差的导数等于导数的和/差：

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

常数的导数是$$$0$$$:

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

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

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

化简：

$$- \frac{1}{1 - x} = \frac{1}{x - 1}$$

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