Ableitung von $$$3 e^{- 4 r} \sin{\left(3 \theta \right)}$$$ nach $$$r$$$
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Ihre Eingabe
Bestimme $$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right)$$$.
Lösung
Wende die Konstantenfaktorregel $$$\frac{d}{dr} \left(c f{\left(r \right)}\right) = c \frac{d}{dr} \left(f{\left(r \right)}\right)$$$ mit $$$c = 3 \sin{\left(3 \theta \right)}$$$ und $$$f{\left(r \right)} = e^{- 4 r}$$$ an:
$${\color{red}\left(\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right)\right)} = {\color{red}\left(3 \sin{\left(3 \theta \right)} \frac{d}{dr} \left(e^{- 4 r}\right)\right)}$$Die Funktion $$$e^{- 4 r}$$$ ist die Komposition $$$f{\left(g{\left(r \right)} \right)}$$$ der beiden Funktionen $$$f{\left(u \right)} = e^{u}$$$ und $$$g{\left(r \right)} = - 4 r$$$.
Wende die Kettenregel $$$\frac{d}{dr} \left(f{\left(g{\left(r \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dr} \left(g{\left(r \right)}\right)$$$ an:
$$3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(e^{- 4 r}\right)\right)} = 3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dr} \left(- 4 r\right)\right)}$$Die Ableitung der Exponentialfunktion ist $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$$3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dr} \left(- 4 r\right) = 3 \sin{\left(3 \theta \right)} {\color{red}\left(e^{u}\right)} \frac{d}{dr} \left(- 4 r\right)$$Zurück zur ursprünglichen Variable:
$$3 e^{{\color{red}\left(u\right)}} \sin{\left(3 \theta \right)} \frac{d}{dr} \left(- 4 r\right) = 3 e^{{\color{red}\left(- 4 r\right)}} \sin{\left(3 \theta \right)} \frac{d}{dr} \left(- 4 r\right)$$Wende die Konstantenfaktorregel $$$\frac{d}{dr} \left(c f{\left(r \right)}\right) = c \frac{d}{dr} \left(f{\left(r \right)}\right)$$$ mit $$$c = -4$$$ und $$$f{\left(r \right)} = r$$$ an:
$$3 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(- 4 r\right)\right)} = 3 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(- 4 \frac{d}{dr} \left(r\right)\right)}$$Wenden Sie die Potenzregel $$$\frac{d}{dr} \left(r^{n}\right) = n r^{n - 1}$$$ mit $$$n = 1$$$ an, mit anderen Worten, $$$\frac{d}{dr} \left(r\right) = 1$$$:
$$- 12 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(r\right)\right)} = - 12 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(1\right)}$$Somit gilt $$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right) = - 12 e^{- 4 r} \sin{\left(3 \theta \right)}$$$.
Antwort
$$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right) = - 12 e^{- 4 r} \sin{\left(3 \theta \right)}$$$A