Afgeleide van $$$x^{2} + \sqrt{2} x + 1$$$

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

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

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

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

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

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

Pas de regel van de constante factor $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ toe met $$$c = \sqrt{2}$$$ en $$$f{\left(x \right)} = x$$$:

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

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

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

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