Derivative of $$$\ln\left(1 - x\right)$$$

The function $$$\ln\left(1 - x\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = 1 - x$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Return to the old variable:

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

The derivative of a sum/difference is the sum/difference of derivatives:

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

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

The derivative of a constant is $$$0$$$:

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

Simplify:

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

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