Funktion $$$\ln\left(1 - x\right)$$$ derivaatta

Funktio $$$\ln\left(1 - x\right)$$$ on kahden funktion $$$f{\left(u \right)} = \ln\left(u\right)$$$ ja $$$g{\left(x \right)} = 1 - x$$$ yhdistelmä $$$f{\left(g{\left(x \right)} \right)}$$$.

Sovella ketjusääntöä $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

Luonnollisen logaritmin derivaatta on $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Palaa alkuperäiseen muuttujaan:

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

Summan/erotuksen derivaatta on derivaattojen summa/erotus:

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

Vakion derivaatta on $$$0$$$:

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

Sovella potenssisääntöä $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ käyttäen $$$n = 1$$$, toisin sanoen, $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

Sievennä:

$$- \frac{1}{1 - x} = \frac{1}{x - 1}$$

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