Afgeleide van $$$x^{2} + 6 x + 25$$$

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

$${\color{red}\left(\frac{d}{dx} \left(x^{2} + 6 x + 25\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(6 x\right) + \frac{d}{dx} \left(25\right)\right)}$$

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

$${\color{red}\left(\frac{d}{dx} \left(25\right)\right)} + \frac{d}{dx} \left(6 x\right) + \frac{d}{dx} \left(x^{2}\right) = {\color{red}\left(0\right)} + \frac{d}{dx} \left(6 x\right) + \frac{d}{dx} \left(x^{2}\right)$$

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 $$$c = 6$$$ en $$$f{\left(x \right)} = x$$$:

$${\color{red}\left(\frac{d}{dx} \left(6 x\right)\right)} + \frac{d}{dx} \left(x^{2}\right) = {\color{red}\left(6 \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(x^{2}\right)$$

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

$$6 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(x^{2}\right) = 6 {\color{red}\left(1\right)} + \frac{d}{dx} \left(x^{2}\right)$$

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

$${\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + 6 = {\color{red}\left(2 x\right)} + 6$$

Dus, $$$\frac{d}{dx} \left(x^{2} + 6 x + 25\right) = 2 x + 6$$$.