Ableitung von $$$\frac{1}{\sqrt{x^{2} - 3 x + 9}}$$$

Die Funktion $$$\frac{1}{\sqrt{x^{2} - 3 x + 9}}$$$ ist die Komposition $$$f{\left(g{\left(x \right)} \right)}$$$ der beiden Funktionen $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$ und $$$g{\left(x \right)} = x^{2} - 3 x + 9$$$.

Wende die Kettenregel $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ an:

$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{\sqrt{x^{2} - 3 x + 9}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right) \frac{d}{dx} \left(x^{2} - 3 x + 9\right)\right)}$$

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

$${\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right)\right)} \frac{d}{dx} \left(x^{2} - 3 x + 9\right) = {\color{red}\left(- \frac{1}{2 u^{\frac{3}{2}}}\right)} \frac{d}{dx} \left(x^{2} - 3 x + 9\right)$$

Zurück zur ursprünglichen Variable:

$$- \frac{\frac{d}{dx} \left(x^{2} - 3 x + 9\right)}{2 {\color{red}\left(u\right)}^{\frac{3}{2}}} = - \frac{\frac{d}{dx} \left(x^{2} - 3 x + 9\right)}{2 {\color{red}\left(x^{2} - 3 x + 9\right)}^{\frac{3}{2}}}$$

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

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2} - 3 x + 9\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(3 x\right) + \frac{d}{dx} \left(9\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

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

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(9\right)\right)} - \frac{d}{dx} \left(3 x\right) + \frac{d}{dx} \left(x^{2}\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{{\color{red}\left(0\right)} - \frac{d}{dx} \left(3 x\right) + \frac{d}{dx} \left(x^{2}\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

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

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} - \frac{d}{dx} \left(3 x\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{{\color{red}\left(2 x\right)} - \frac{d}{dx} \left(3 x\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

Wende die Konstantenfaktorregel $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ mit $$$c = 3$$$ und $$$f{\left(x \right)} = x$$$ an:

$$- \frac{2 x - {\color{red}\left(\frac{d}{dx} \left(3 x\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{2 x - {\color{red}\left(3 \frac{d}{dx} \left(x\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

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

$$- \frac{2 x - 3 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{2 x - 3 {\color{red}\left(1\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

Vereinfachen:

$$- \frac{2 x - 3}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = \frac{\frac{3}{2} - x}{\left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

Somit gilt $$$\frac{d}{dx} \left(\frac{1}{\sqrt{x^{2} - 3 x + 9}}\right) = \frac{\frac{3}{2} - x}{\left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$$.