$$$\ln\left(-1 + \frac{1}{x}\right)$$$ 的導數

函數 $$$\ln\left(-1 + \frac{1}{x}\right)$$$ 是兩個函數 $$$f{\left(u \right)} = \ln\left(u\right)$$$ 與 $$$g{\left(x \right)} = -1 + \frac{1}{x}$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。

應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

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

自然對數的導數為 $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$：

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

返回原變數：

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

和/差的導數等於導數的和/差：

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

套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = -1$$$：

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

常數的導數為$$$0$$$：

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

化簡：

$$- \frac{1}{x^{2} \left(-1 + \frac{1}{x}\right)} = \frac{1}{x \left(x - 1\right)}$$

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