$$$t \left(t - 1\right)$$$ 的導數

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 $$$f{\left(t \right)} = t$$$ 和 $$$g{\left(t \right)} = t - 1$$$：

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

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

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

和/差的導數等於導數的和/差：

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

常數的導數為$$$0$$$：

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

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

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

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