$$$t^{2} - 1$$$的导数
您的输入
求$$$\frac{d}{dt} \left(t^{2} - 1\right)$$$。
解答
和/差的导数等于导数的和/差：
$${\color{red}\left(\frac{d}{dt} \left(t^{2} - 1\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(t^{2}\right) - \frac{d}{dt} \left(1\right)\right)}$$应用幂次法则 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，其中 $$$n = 2$$$:
$${\color{red}\left(\frac{d}{dt} \left(t^{2}\right)\right)} - \frac{d}{dt} \left(1\right) = {\color{red}\left(2 t\right)} - \frac{d}{dt} \left(1\right)$$常数的导数是$$$0$$$:
$$2 t - {\color{red}\left(\frac{d}{dt} \left(1\right)\right)} = 2 t - {\color{red}\left(0\right)}$$因此，$$$\frac{d}{dt} \left(t^{2} - 1\right) = 2 t$$$。
答案
$$$\frac{d}{dt} \left(t^{2} - 1\right) = 2 t$$$A