Afgeleide van $$$\ln\left(\frac{x + 1}{1 - x}\right)$$$

De functie $$$\ln\left(\frac{x + 1}{1 - x}\right)$$$ is de samenstelling $$$f{\left(g{\left(x \right)} \right)}$$$ van twee functies $$$f{\left(u \right)} = \ln\left(u\right)$$$ en $$$g{\left(x \right)} = \frac{x + 1}{1 - x}$$$.

Pas de kettingregel $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ toe:

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

De afgeleide van de natuurlijke logaritme is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Keer terug naar de oorspronkelijke variabele:

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

Pas de quotiëntregel $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$ toe op $$$f{\left(x \right)} = x + 1$$$ en $$$g{\left(x \right)} = 1 - x$$$:

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

De afgeleide van een som/verschil is de som/het verschil van de afgeleiden:

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

De afgeleide van een constante is $$$0$$$:

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

De afgeleide van een som/verschil is de som/het verschil van de afgeleiden:

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

De afgeleide van een constante is $$$0$$$:

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

Pas de machtsregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ toe met $$$n = 1$$$, met andere woorden, $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

Vereenvoudig:

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

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