Ableitung von $$$- \frac{2 x}{x^{2} + 1}$$$

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

Wende die Konstantenfaktorregel $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ mit $$$c = -2$$$ und $$$f{\left(x \right)} = \frac{x}{x^{2} + 1}$$$ an:

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

Wende die Quotientenregel $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$ auf $$$f{\left(x \right)} = x$$$ und $$$g{\left(x \right)} = x^{2} + 1$$$ an:

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

Die Ableitung einer Summe/Differenz ist die Summe/Differenz der Ableitungen:

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

Wende die Potenzregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ mit $$$n = 2$$$ an:

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

Die Ableitung einer Konstante ist $$$0$$$:

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

Wenden Sie die Potenzregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ mit $$$n = 1$$$ an, mit anderen Worten, $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

Vereinfachen:

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

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