Derivative of $$$\sqrt{2} t - \sqrt{-3 + \sqrt{5}}$$$

The derivative of a sum/difference is the sum/difference of derivatives:

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

Apply the constant multiple rule $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ with $$$c = \sqrt{2}$$$ and $$$f{\left(t \right)} = t$$$:

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

Apply the power rule $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dt} \left(t\right) = 1$$$:

$$\sqrt{2} {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} - \frac{d}{dt} \left(\sqrt{-3 + \sqrt{5}}\right) = \sqrt{2} {\color{red}\left(1\right)} - \frac{d}{dt} \left(\sqrt{-3 + \sqrt{5}}\right)$$

The derivative of a constant is $$$0$$$:

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

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