$$$\ln\left(\frac{x + 1}{1 - x}\right)$$$의 도함수

함수 $$$\ln\left(\frac{x + 1}{1 - x}\right)$$$는 두 함수 $$$f{\left(u \right)} = \ln\left(u\right)$$$와 $$$g{\left(x \right)} = \frac{x + 1}{1 - x}$$$의 합성함수 $$$f{\left(g{\left(x \right)} \right)}$$$이다.

연쇄법칙 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$을(를) 적용하십시오:

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

자연로그 함수의 도함수는 $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

역치환:

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

$$$f{\left(x \right)} = x + 1$$$와 $$$g{\left(x \right)} = 1 - x$$$에 대해 몫의 미분법칙 $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$을 적용하십시오:

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

합/차의 도함수는 도함수들의 합/차이다:

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

상수의 도함수는 $$$0$$$입니다:

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

합/차의 도함수는 도함수들의 합/차이다:

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

상수의 도함수는 $$$0$$$입니다:

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

멱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 1$$$에 대해 적용하면, 즉 $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

간단히 하시오:

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

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