$$$\cos{\left(t \right)} - \cos{\left(2 t \right)}$$$ 的導數

和/差的導數等於導數的和/差：

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

餘弦函數的導數為 $$$\frac{d}{dt} \left(\cos{\left(t \right)}\right) = - \sin{\left(t \right)}$$$：

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

函數 $$$\cos{\left(2 t \right)}$$$ 是兩個函數 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ 與 $$$g{\left(t \right)} = 2 t$$$ 之複合 $$$f{\left(g{\left(t \right)} \right)}$$$。

應用鏈式法則 $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$：

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

餘弦函數的導數為 $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$：

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

返回原變數：

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

套用常數倍法則 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$，使用 $$$c = 2$$$ 與 $$$f{\left(t \right)} = t$$$：

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

套用冪次法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，取 $$$n = 1$$$，也就是 $$$\frac{d}{dt} \left(t\right) = 1$$$：

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

化簡：

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

因此，$$$\frac{d}{dt} \left(\cos{\left(t \right)} - \cos{\left(2 t \right)}\right) = \left(4 \cos{\left(t \right)} - 1\right) \sin{\left(t \right)}$$$。