Ableitung von $$$e^{t} \cos{\left(t \right)}$$$
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Ihre Eingabe
Bestimme $$$\frac{d}{dt} \left(e^{t} \cos{\left(t \right)}\right)$$$.
Lösung
Wende die Produktregel $$$\frac{d}{dt} \left(f{\left(t \right)} g{\left(t \right)}\right) = \frac{d}{dt} \left(f{\left(t \right)}\right) g{\left(t \right)} + f{\left(t \right)} \frac{d}{dt} \left(g{\left(t \right)}\right)$$$ mit $$$f{\left(t \right)} = \cos{\left(t \right)}$$$ und $$$g{\left(t \right)} = e^{t}$$$ an:
$${\color{red}\left(\frac{d}{dt} \left(e^{t} \cos{\left(t \right)}\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(\cos{\left(t \right)}\right) e^{t} + \cos{\left(t \right)} \frac{d}{dt} \left(e^{t}\right)\right)}$$Die Ableitung des Kosinus ist $$$\frac{d}{dt} \left(\cos{\left(t \right)}\right) = - \sin{\left(t \right)}$$$:
$$e^{t} {\color{red}\left(\frac{d}{dt} \left(\cos{\left(t \right)}\right)\right)} + \cos{\left(t \right)} \frac{d}{dt} \left(e^{t}\right) = e^{t} {\color{red}\left(- \sin{\left(t \right)}\right)} + \cos{\left(t \right)} \frac{d}{dt} \left(e^{t}\right)$$Die Ableitung der Exponentialfunktion ist $$$\frac{d}{dt} \left(e^{t}\right) = e^{t}$$$:
$$- e^{t} \sin{\left(t \right)} + \cos{\left(t \right)} {\color{red}\left(\frac{d}{dt} \left(e^{t}\right)\right)} = - e^{t} \sin{\left(t \right)} + \cos{\left(t \right)} {\color{red}\left(e^{t}\right)}$$Vereinfachen:
$$- e^{t} \sin{\left(t \right)} + e^{t} \cos{\left(t \right)} = \sqrt{2} e^{t} \cos{\left(t + \frac{\pi}{4} \right)}$$Somit gilt $$$\frac{d}{dt} \left(e^{t} \cos{\left(t \right)}\right) = \sqrt{2} e^{t} \cos{\left(t + \frac{\pi}{4} \right)}$$$.
Antwort
$$$\frac{d}{dt} \left(e^{t} \cos{\left(t \right)}\right) = \sqrt{2} e^{t} \cos{\left(t + \frac{\pi}{4} \right)}$$$A