Afgeleide van $$$\ln\left(x + \sqrt{x^{2} + 1}\right)$$$

Bepaal $$$\frac{d}{dx} \left(\ln\left(x + \sqrt{x^{2} + 1}\right)\right)$$$.

De functie $$$\ln\left(x + \sqrt{x^{2} + 1}\right)$$$ is de samenstelling $$$f{\left(g{\left(x \right)} \right)}$$$ van twee functies $$$f{\left(u \right)} = \ln\left(u\right)$$$ en $$$g{\left(x \right)} = x + \sqrt{x^{2} + 1}$$$.

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:

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

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

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

Keer terug naar de oorspronkelijke variabele:

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

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

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

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

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

De functie $$$\sqrt{x^{2} + 1}$$$ is de samenstelling $$$f{\left(g{\left(x \right)} \right)}$$$ van twee functies $$$f{\left(u \right)} = \sqrt{u}$$$ en $$$g{\left(x \right)} = x^{2} + 1$$$.

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:

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

Pas de machtsregel $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ toe met $$$n = \frac{1}{2}$$$:

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

Keer terug naar de oorspronkelijke variabele:

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

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

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

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

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

Pas de machtsregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ toe met $$$n = 2$$$:

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

Vereenvoudig:

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

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