Derivatan av $$$\frac{1}{x^{2} + 1}$$$

Funktionen $$$\frac{1}{x^{2} + 1}$$$ är sammansättningen $$$f{\left(g{\left(x \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = \frac{1}{u}$$$ och $$$g{\left(x \right)} = x^{2} + 1$$$.

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

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

Tillämpa potensregeln $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ med $$$n = -1$$$:

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

Återgå till den ursprungliga variabeln:

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

Derivatan av en summa/differens är summan/differensen av derivatorna:

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

Derivatan av en konstant är $$$0$$$:

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

Tillämpa potensregeln $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ med $$$n = 2$$$:

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

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