$$$t \left(t - 1\right)$$$的导数

对 $$$f{\left(t \right)} = t$$$ 和 $$$g{\left(t \right)} = t - 1$$$ 应用乘积法则 $$$\frac{d}{dt} \left(f{\left(t \right)} g{\left(t \right)}\right) = \frac{d}{dt} \left(f{\left(t \right)}\right) g{\left(t \right)} + f{\left(t \right)} \frac{d}{dt} \left(g{\left(t \right)}\right)$$$:

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

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

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

和/差的导数等于导数的和/差：

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

常数的导数是$$$0$$$:

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

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

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

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