$$$2 t - 1 + \frac{1}{t}$$$的导数

和/差的导数等于导数的和/差：

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

常数的导数是$$$0$$$:

$$- {\color{red}\left(\frac{d}{dt} \left(1\right)\right)} + \frac{d}{dt} \left(\frac{1}{t}\right) + \frac{d}{dt} \left(2 t\right) = - {\color{red}\left(0\right)} + \frac{d}{dt} \left(\frac{1}{t}\right) + \frac{d}{dt} \left(2 t\right)$$

对 $$$c = 2$$$ 和 $$$f{\left(t \right)} = t$$$ 应用常数倍法则 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$：

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

应用幂法则 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dt} \left(t\right) = 1$$$：

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

应用幂次法则 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，其中 $$$n = -1$$$:

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

因此，$$$\frac{d}{dt} \left(2 t - 1 + \frac{1}{t}\right) = 2 - \frac{1}{t^{2}}$$$。