Derivada de $$$\ln\left(x^{2} - 2 x + 1\right)$$$

A função $$$\ln\left(x^{2} - 2 x + 1\right)$$$ é a composição $$$f{\left(g{\left(x \right)} \right)}$$$ de duas funções $$$f{\left(u \right)} = \ln\left(u\right)$$$ e $$$g{\left(x \right)} = x^{2} - 2 x + 1$$$.

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

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

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

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

Retorne à variável original:

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

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

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

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

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

Aplique a regra da constante multiplicativa $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ com $$$c = 2$$$ e $$$f{\left(x \right)} = x$$$:

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

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

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

Aplique a regra da potência $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ com $$$n = 2$$$:

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

Simplifique:

$$\frac{2 x - 2}{x^{2} - 2 x + 1} = \frac{2}{x - 1}$$

Logo, $$$\frac{d}{dx} \left(\ln\left(x^{2} - 2 x + 1\right)\right) = \frac{2}{x - 1}$$$.