Ableitung von $$$\cos{\left(t \right)} - \cos{\left(2 t \right)}$$$

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

$${\color{red}\left(\frac{d}{dt} \left(\cos{\left(t \right)} - \cos{\left(2 t \right)}\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(\cos{\left(t \right)}\right) - \frac{d}{dt} \left(\cos{\left(2 t \right)}\right)\right)}$$

Die Ableitung des Kosinus ist $$$\frac{d}{dt} \left(\cos{\left(t \right)}\right) = - \sin{\left(t \right)}$$$:

$${\color{red}\left(\frac{d}{dt} \left(\cos{\left(t \right)}\right)\right)} - \frac{d}{dt} \left(\cos{\left(2 t \right)}\right) = {\color{red}\left(- \sin{\left(t \right)}\right)} - \frac{d}{dt} \left(\cos{\left(2 t \right)}\right)$$

Die Funktion $$$\cos{\left(2 t \right)}$$$ ist die Komposition $$$f{\left(g{\left(t \right)} \right)}$$$ der beiden Funktionen $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ und $$$g{\left(t \right)} = 2 t$$$.

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

$$- \sin{\left(t \right)} - {\color{red}\left(\frac{d}{dt} \left(\cos{\left(2 t \right)}\right)\right)} = - \sin{\left(t \right)} - {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{dt} \left(2 t\right)\right)}$$

Die Ableitung des Kosinus ist $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:

$$- \sin{\left(t \right)} - {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{dt} \left(2 t\right) = - \sin{\left(t \right)} - {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{dt} \left(2 t\right)$$

Zurück zur ursprünglichen Variable:

$$- \sin{\left(t \right)} + \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{dt} \left(2 t\right) = - \sin{\left(t \right)} + \sin{\left({\color{red}\left(2 t\right)} \right)} \frac{d}{dt} \left(2 t\right)$$

Wende die Konstantenfaktorregel $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ mit $$$c = 2$$$ und $$$f{\left(t \right)} = t$$$ an:

$$- \sin{\left(t \right)} + \sin{\left(2 t \right)} {\color{red}\left(\frac{d}{dt} \left(2 t\right)\right)} = - \sin{\left(t \right)} + \sin{\left(2 t \right)} {\color{red}\left(2 \frac{d}{dt} \left(t\right)\right)}$$

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

$$- \sin{\left(t \right)} + 2 \sin{\left(2 t \right)} {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} = - \sin{\left(t \right)} + 2 \sin{\left(2 t \right)} {\color{red}\left(1\right)}$$

Vereinfachen:

$$- \sin{\left(t \right)} + 2 \sin{\left(2 t \right)} = \left(4 \cos{\left(t \right)} - 1\right) \sin{\left(t \right)}$$

Somit gilt $$$\frac{d}{dt} \left(\cos{\left(t \right)} - \cos{\left(2 t \right)}\right) = \left(4 \cos{\left(t \right)} - 1\right) \sin{\left(t \right)}$$$.