Derivative of $$$a t - b t$$$ with respect to $$$t$$$
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Find $$$\frac{d}{dt} \left(a t - b t\right)$$$.
Solution
The derivative of a sum/difference is the sum/difference of derivatives:
$${\color{red}\left(\frac{d}{dt} \left(a t - b t\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(a t\right) - \frac{d}{dt} \left(b t\right)\right)}$$Apply the constant multiple rule $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ with $$$c = b$$$ and $$$f{\left(t \right)} = t$$$:
$$- {\color{red}\left(\frac{d}{dt} \left(b t\right)\right)} + \frac{d}{dt} \left(a t\right) = - {\color{red}\left(b \frac{d}{dt} \left(t\right)\right)} + \frac{d}{dt} \left(a t\right)$$Apply the power rule $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dt} \left(t\right) = 1$$$:
$$- b {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} + \frac{d}{dt} \left(a t\right) = - b {\color{red}\left(1\right)} + \frac{d}{dt} \left(a t\right)$$Apply the constant multiple rule $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ with $$$c = a$$$ and $$$f{\left(t \right)} = t$$$:
$$- b + {\color{red}\left(\frac{d}{dt} \left(a t\right)\right)} = - b + {\color{red}\left(a \frac{d}{dt} \left(t\right)\right)}$$Apply the power rule $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dt} \left(t\right) = 1$$$:
$$a {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} - b = a {\color{red}\left(1\right)} - b$$Thus, $$$\frac{d}{dt} \left(a t - b t\right) = a - b$$$.
Answer
$$$\frac{d}{dt} \left(a t - b t\right) = a - b$$$A