$$$- 2 e^{t} \sin{\left(t \right)}$$$ 的導數

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 $$$f{\left(t \right)} = e^{t} \sin{\left(t \right)}$$$：

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

將乘積法則 $$$\frac{d}{dt} \left(f{\left(t \right)} g{\left(t \right)}\right) = \frac{d}{dt} \left(f{\left(t \right)}\right) g{\left(t \right)} + f{\left(t \right)} \frac{d}{dt} \left(g{\left(t \right)}\right)$$$ 應用於 $$$f{\left(t \right)} = e^{t}$$$ 和 $$$g{\left(t \right)} = \sin{\left(t \right)}$$$：

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

指數函數的導數為 $$$\frac{d}{dt} \left(e^{t}\right) = e^{t}$$$：

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

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

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

化簡：

$$- 2 e^{t} \sin{\left(t \right)} - 2 e^{t} \cos{\left(t \right)} = - 2 \sqrt{2} e^{t} \sin{\left(t + \frac{\pi}{4} \right)}$$

因此，$$$\frac{d}{dt} \left(- 2 e^{t} \sin{\left(t \right)}\right) = - 2 \sqrt{2} e^{t} \sin{\left(t + \frac{\pi}{4} \right)}$$$。