$$$a t - b t$$$ 关于 $$$t$$$ 的导数
您的输入
求$$$\frac{d}{dt} \left(a t - b t\right)$$$。
解答
和/差的导数等于导数的和/差:
$${\color{red}\left(\frac{d}{dt} \left(a t - b t\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(a t\right) - \frac{d}{dt} \left(b t\right)\right)}$$对 $$$c = b$$$ 和 $$$f{\left(t \right)} = t$$$ 应用常数倍法则 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$:
$$- {\color{red}\left(\frac{d}{dt} \left(b t\right)\right)} + \frac{d}{dt} \left(a t\right) = - {\color{red}\left(b \frac{d}{dt} \left(t\right)\right)} + \frac{d}{dt} \left(a t\right)$$应用幂法则 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$,取 $$$n = 1$$$,也就是说,$$$\frac{d}{dt} \left(t\right) = 1$$$:
$$- b {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} + \frac{d}{dt} \left(a t\right) = - b {\color{red}\left(1\right)} + \frac{d}{dt} \left(a t\right)$$对 $$$c = a$$$ 和 $$$f{\left(t \right)} = t$$$ 应用常数倍法则 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$:
$$- b + {\color{red}\left(\frac{d}{dt} \left(a t\right)\right)} = - b + {\color{red}\left(a \frac{d}{dt} \left(t\right)\right)}$$应用幂法则 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$,取 $$$n = 1$$$,也就是说,$$$\frac{d}{dt} \left(t\right) = 1$$$:
$$a {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} - b = a {\color{red}\left(1\right)} - b$$因此,$$$\frac{d}{dt} \left(a t - b t\right) = a - b$$$。
答案
$$$\frac{d}{dt} \left(a t - b t\right) = a - b$$$A