Turunan dari $$$2 t - 1 + \frac{1}{t}$$$

Turunan dari jumlah/selisih adalah jumlah/selisih dari turunan:

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

Turunan dari suatu konstanta adalah $$$0$$$:

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

Terapkan aturan kelipatan konstanta $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ dengan $$$c = 2$$$ dan $$$f{\left(t \right)} = t$$$:

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

Terapkan aturan pangkat $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ dengan $$$n = 1$$$, dengan kata lain, $$$\frac{d}{dt} \left(t\right) = 1$$$:

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

Terapkan aturan pangkat $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ dengan $$$n = -1$$$:

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

Dengan demikian, $$$\frac{d}{dt} \left(2 t - 1 + \frac{1}{t}\right) = 2 - \frac{1}{t^{2}}$$$.