Afgeleide van $$$\frac{1}{u^{2} + 1}$$$

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

Pas de kettingregel $$$\frac{d}{du} \left(f{\left(g{\left(u \right)} \right)}\right) = \frac{d}{dv} \left(f{\left(v \right)}\right) \frac{d}{du} \left(g{\left(u \right)}\right)$$$ toe:

$${\color{red}\left(\frac{d}{du} \left(\frac{1}{u^{2} + 1}\right)\right)} = {\color{red}\left(\frac{d}{dv} \left(\frac{1}{v}\right) \frac{d}{du} \left(u^{2} + 1\right)\right)}$$

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

$${\color{red}\left(\frac{d}{dv} \left(\frac{1}{v}\right)\right)} \frac{d}{du} \left(u^{2} + 1\right) = {\color{red}\left(- \frac{1}{v^{2}}\right)} \frac{d}{du} \left(u^{2} + 1\right)$$

Keer terug naar de oorspronkelijke variabele:

$$- \frac{\frac{d}{du} \left(u^{2} + 1\right)}{{\color{red}\left(v\right)}^{2}} = - \frac{\frac{d}{du} \left(u^{2} + 1\right)}{{\color{red}\left(u^{2} + 1\right)}^{2}}$$

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

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

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

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

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

$$- \frac{2 u + {\color{red}\left(\frac{d}{du} \left(1\right)\right)}}{\left(u^{2} + 1\right)^{2}} = - \frac{2 u + {\color{red}\left(0\right)}}{\left(u^{2} + 1\right)^{2}}$$

Dus, $$$\frac{d}{du} \left(\frac{1}{u^{2} + 1}\right) = - \frac{2 u}{\left(u^{2} + 1\right)^{2}}$$$.