$$$\sqrt{2} t - \sqrt{-3 + \sqrt{5}}$$$ 的導數

和/差的導數等於導數的和/差：

$${\color{red}\left(\frac{d}{dt} \left(\sqrt{2} t - \sqrt{-3 + \sqrt{5}}\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(\sqrt{2} t\right) - \frac{d}{dt} \left(\sqrt{-3 + \sqrt{5}}\right)\right)}$$

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 $$$c = \sqrt{2}$$$ 與 $$$f{\left(t \right)} = t$$$：

$${\color{red}\left(\frac{d}{dt} \left(\sqrt{2} t\right)\right)} - \frac{d}{dt} \left(\sqrt{-3 + \sqrt{5}}\right) = {\color{red}\left(\sqrt{2} \frac{d}{dt} \left(t\right)\right)} - \frac{d}{dt} \left(\sqrt{-3 + \sqrt{5}}\right)$$

常數的導數為$$$0$$$：

$$- {\color{red}\left(\frac{d}{dt} \left(\sqrt{-3 + \sqrt{5}}\right)\right)} + \sqrt{2} \frac{d}{dt} \left(t\right) = - {\color{red}\left(0\right)} + \sqrt{2} \frac{d}{dt} \left(t\right)$$

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

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

因此，$$$\frac{d}{dt} \left(\sqrt{2} t - \sqrt{-3 + \sqrt{5}}\right) = \sqrt{2}$$$。