Funktion $$$\ln\left(x^{2} + 1\right)$$$ derivaatta

Funktio $$$\ln\left(x^{2} + 1\right)$$$ on kahden funktion $$$f{\left(u \right)} = \ln\left(u\right)$$$ ja $$$g{\left(x \right)} = x^{2} + 1$$$ yhdistelmä $$$f{\left(g{\left(x \right)} \right)}$$$.

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

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

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

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

Palaa alkuperäiseen muuttujaan:

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

Summan/erotuksen derivaatta on derivaattojen summa/erotus:

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

Vakion derivaatta on $$$0$$$:

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

Sovella potenssisääntöä $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$, kun $$$n = 2$$$:

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

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