Lösung zur Ableitung der Formel
Eingabefunktion
$$\frac{x^{2} - 3\,x + 2}{x + 4}$$
Start der Ableitung
$$\frac{d}{dx}{\left(\frac{x^{2} - 3\,x + 2}{x + 4}\right)}$$
Schritt 1 — Quotientenregel
Regel
$$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^{2}}$$
Mit:
- $u = x^{2} - 3\,x + 2$
- $v = x + 4$
Aktueller Ausdruck
$$\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(\frac{x^{2} - 3\,x + 2}{x + 4}\right)}\,}}}$$
Nach Quotientenregel
$$\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{\frac{d}{dx}{\left(x^{2} - 3\,x + 2\right)}\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}\,}}}$$
Schritt 2 — Summen-/Differenzenregel
Regel
$$\frac{d}{dx}(u \pm v \pm \dots) = \frac{d}{dx}(u) \pm \frac{d}{dx}(v) \pm \dots$$
Mit:
- $x^{2}$
- $-3\,x$
- $2$
Aktueller Ausdruck
$$\frac{\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x^{2} - 3\,x + 2\right)}\,}}}\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Nach Summen-/Differenzenregel
$$\frac{\left(\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x^{2}\right)} - \frac{d}{dx}{\left(3\,x\right)} + \frac{d}{dx}{\left(2\right)}\,}}}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Schritt 3 — Potenzregel (Spezialfall)
Regel
$$\frac{d}{dx}\left(u^{c}\right)=c\,u^{c-1}$$
Mit:
- $u = x$
- $c = 2$
Aktueller Ausdruck
$$\frac{\left(\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x^{2}\right)}\,}}} - \frac{d}{dx}{\left(3\,x\right)} + \frac{d}{dx}{\left(2\right)}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Nach Potenzregel (Spezialfall)
$$\frac{\left(\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,2\,x^{2 - 1}\,}}} - \frac{d}{dx}{\left(3\,x\right)} + \frac{d}{dx}{\left(2\right)}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Vereinfacht
$$\frac{\left(\bbox[lightgreen, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,2\,x\,}}} - \frac{d}{dx}{\left(3\,x\right)} + \frac{d}{dx}{\left(2\right)}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Schritt 4 — Faktorregel
Regel
$$\frac{d}{dx}(c\,u)=c\,u'$$
Mit:
- $c = 3$
- $u = x$
Aktueller Ausdruck
$$\frac{\left(2\,x - \bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(3\,x\right)}\,}}} + \frac{d}{dx}{\left(2\right)}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Nach Faktorregel
$$\frac{\left(2\,x - \bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,3\,\frac{d}{dx}{\left(x\right)}\,}}} + \frac{d}{dx}{\left(2\right)}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Schritt 5 — Ableitung der Variablen
Regel
$$\frac{d}{dx}(x)=1$$
Aktueller Ausdruck
$$\frac{\left(2\,x - 3\,\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x\right)}\,}}} + \frac{d}{dx}{\left(2\right)}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Nach Ableitung der Variablen
$$\frac{\left(2\,x - 3 \cdot \bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,1\,}}} + \frac{d}{dx}{\left(2\right)}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Vereinfacht
$$\frac{\left(2\,x - \bbox[lightgreen, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,3\,}}} + \frac{d}{dx}{\left(2\right)}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Schritt 6 — Konstantenregel
Regel
$$\frac{d}{dx}(c)=0$$
Mit:
- $c = 2$
Aktueller Ausdruck
$$\frac{\left(2\,x - 3 + \bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(2\right)}\,}}}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Nach Konstantenregel
$$\frac{\left(2\,x - 3 + \bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,0\,}}}\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Vereinfacht
$$\frac{\bbox[lightgreen, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\left(2\,x - 3\right)\,\left(x + 4\right)\,}}} - \left(x^{2} - 3\,x + 2\right)\,\frac{d}{dx}{\left(x + 4\right)}}{\left(x + 4\right)^{2}}$$
Schritt 7 — Summen-/Differenzenregel
Regel
$$\frac{d}{dx}(u \pm v \pm \dots) = \frac{d}{dx}(u) \pm \frac{d}{dx}(v) \pm \dots$$
Mit:
- $u = x$
- $v = 4$
Aktueller Ausdruck
$$\frac{\left(2\,x - 3\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)\,\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x + 4\right)}\,}}}}{\left(x + 4\right)^{2}}$$
Nach Summen-/Differenzenregel
$$\frac{\left(2\,x - 3\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right) \cdot \left(\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x\right)} + \frac{d}{dx}{\left(4\right)}\,}}}\right)}{\left(x + 4\right)^{2}}$$
Schritt 8 — Ableitung der Variablen
Regel
$$\frac{d}{dx}(x)=1$$
Aktueller Ausdruck
$$\frac{\left(2\,x - 3\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right) \cdot \left(\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(x\right)}\,}}} + \frac{d}{dx}{\left(4\right)}\right)}{\left(x + 4\right)^{2}}$$
Nach Ableitung der Variablen
$$\frac{\left(2\,x - 3\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right) \cdot \left(\bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,1\,}}} + \frac{d}{dx}{\left(4\right)}\right)}{\left(x + 4\right)^{2}}$$
Schritt 9 — Konstantenregel
Regel
$$\frac{d}{dx}(c)=0$$
Mit:
- $c = 4$
Aktueller Ausdruck
$$\frac{\left(2\,x - 3\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right) \cdot \left(1 + \bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left(4\right)}\,}}}\right)}{\left(x + 4\right)^{2}}$$
Nach Konstantenregel
$$\frac{\left(2\,x - 3\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right) \cdot \left(1 + \bbox[yellow, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,0\,}}}\right)}{\left(x + 4\right)^{2}}$$
Vereinfacht
$$\frac{\left(2\,x - 3\right)\,\left(x + 4\right) - \bbox[lightgreen, padding:0.12em 0.22em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,x^{2} - 3\,x + 2\,}}}}{\left(x + 4\right)^{2}}$$
Ergebnis
$$\frac{\left(2\,x - 3\right)\,\left(x + 4\right) - \left(x^{2} - 3\,x + 2\right)}{\left(x + 4\right)^{2}}$$
Direkt berechnet (Maxima)
Ableitung der Eingabefunktion via Maxima:
$$\frac{2\,x - 3}{x + 4} - \frac{x^{2} - 3\,x + 2}{\left(x + 4\right)^{2}}$$
Abgleich Schrittfolge ↔︎ Maxima:
gleich (true)