Lösung zur Ableitung der Formel
Eingabefunktion
$$\frac{x^{3}}{x^{2} + 1}$$
Start der Ableitung
$$\frac{d}{dx}{\left(\frac{x^{3}}{x^{2} + 1}\right)}$$
Schritt 1 — Quotientenregel
Regel
$$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^{2}}$$
Mit:
- $u = x^{3}$
- $v = x^{2} + 1$
Aktueller Ausdruck
$$\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(\frac{x^{3}}{x^{2} + 1}\right)}\,}}}$$
Nach Quotientenregel
$$\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{\frac{d}{dx}{\left(x^{3}\right)}\left(x^{2} + 1\right) - x^{3}\frac{d}{dx}{\left(x^{2} + 1\right)}}{\left(x^{2} + 1\right)^{2}}\,}}}$$
Schritt 2 — Potenzregel (Spezialfall)
Regel
$$\frac{d}{dx}\left(u^{c}\right)=c\,u^{c-1}$$
Mit:
- $u = x$
- $c = 3$
Aktueller Ausdruck
$$\frac{\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x^{3}\right)}\,}}}\left(x^{2} + 1\right) - x^{3}\frac{d}{dx}{\left(x^{2} + 1\right)}}{\left(x^{2} + 1\right)^{2}}$$
Nach Potenzregel (Spezialfall)
$$\frac{\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,3\,x^{3 - 1}\,}}}\left(x^{2} + 1\right) - x^{3}\frac{d}{dx}{\left(x^{2} + 1\right)}}{\left(x^{2} + 1\right)^{2}}$$
Vereinfacht
$$\frac{3\,\bbox[lightgreen, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,x^{2}\,}}}\left(x^{2} + 1\right) - x^{3}\frac{d}{dx}{\left(x^{2} + 1\right)}}{\left(x^{2} + 1\right)^{2}}$$
Schritt 3 — Summen-/Differenzenregel
Regel
$$\frac{d}{dx}(u \pm v \pm \dots) = \frac{d}{dx}(u) \pm \frac{d}{dx}(v) \pm \dots$$
Mit:
- $u = x^{2}$
- $v = 1$
Aktueller Ausdruck
$$\frac{3\,x^{2}\left(x^{2} + 1\right) - x^{3}\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x^{2} + 1\right)}\,}}}}{\left(x^{2} + 1\right)^{2}}$$
Nach Summen-/Differenzenregel
$$\frac{3\,x^{2}\left(x^{2} + 1\right) - x^{3} \cdot \left(\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x^{2}\right)} + \frac{d}{dx}{\left(1\right)}\,}}}\right)}{\left(x^{2} + 1\right)^{2}}$$
Schritt 4 — Potenzregel (Spezialfall)
Regel
$$\frac{d}{dx}\left(u^{c}\right)=c\,u^{c-1}$$
Mit:
- $u = x$
- $c = 2$
Aktueller Ausdruck
$$\frac{3\,x^{2}\left(x^{2} + 1\right) - x^{3} \cdot \left(\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x^{2}\right)}\,}}} + \frac{d}{dx}{\left(1\right)}\right)}{\left(x^{2} + 1\right)^{2}}$$
Nach Potenzregel (Spezialfall)
$$\frac{3\,x^{2}\left(x^{2} + 1\right) - x^{3} \cdot \left(\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,2\,x^{2 - 1}\,}}} + \frac{d}{dx}{\left(1\right)}\right)}{\left(x^{2} + 1\right)^{2}}$$
Vereinfacht
$$\frac{3\,x^{2}\left(x^{2} + 1\right) - x^{3} \cdot \left(\bbox[lightgreen, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,2\,x\,}}} + \frac{d}{dx}{\left(1\right)}\right)}{\left(x^{2} + 1\right)^{2}}$$
Schritt 5 — Konstantenregel
Regel
$$\frac{d}{dx}(c)=0$$
Mit:
- $c = 1$
Aktueller Ausdruck
$$\frac{3\,x^{2}\left(x^{2} + 1\right) - x^{3} \cdot \left(2\,x + \bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(1\right)}\,}}}\right)}{\left(x^{2} + 1\right)^{2}}$$
Nach Konstantenregel
$$\frac{3\,x^{2}\left(x^{2} + 1\right) - x^{3} \cdot \left(2\,x + \bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,0\,}}}\right)}{\left(x^{2} + 1\right)^{2}}$$
Vereinfacht
$$\frac{3\,x^{2}\left(x^{2} + 1\right) - \bbox[lightgreen, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,x^{3} \cdot 2\,x\,}}}}{\left(x^{2} + 1\right)^{2}}$$
Ergebnis
$$\frac{3\,x^{2}\left(x^{2} + 1\right) - x^{3} \cdot 2\,x}{\left(x^{2} + 1\right)^{2}}$$
Direkt berechnet (Maxima)
Ableitung der Eingabefunktion via Maxima:
$$\frac{3\,x^{2}}{x^{2} + 1} - \frac{2\,x^{4}}{\left(x^{2} + 1\right)^{2}}$$
Abgleich Schrittfolge ↔︎ Maxima:
gleich (true)