Derivative of $$$t \left(t - 1\right)$$$

Apply the product rule $$$\frac{d}{dt} \left(f{\left(t \right)} g{\left(t \right)}\right) = \frac{d}{dt} \left(f{\left(t \right)}\right) g{\left(t \right)} + f{\left(t \right)} \frac{d}{dt} \left(g{\left(t \right)}\right)$$$ with $$$f{\left(t \right)} = t$$$ and $$$g{\left(t \right)} = t - 1$$$:

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

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

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

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

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

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

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

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