Derivada de $$$\ln\left(-1 + \frac{1}{x}\right)$$$

A função $$$\ln\left(-1 + \frac{1}{x}\right)$$$ é a composição $$$f{\left(g{\left(x \right)} \right)}$$$ de duas funções $$$f{\left(u \right)} = \ln\left(u\right)$$$ e $$$g{\left(x \right)} = -1 + \frac{1}{x}$$$.

Aplique a regra da cadeia $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

A derivada do logaritmo natural é $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Retorne à variável original:

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

A derivada de uma soma/diferença é a soma/diferença das derivadas:

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

Aplique a regra da potência $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ com $$$n = -1$$$:

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

A derivada de uma constante é $$$0$$$:

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

Simplifique:

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

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