$$$3 t^{2} - 7$$$的导数
您的输入
求$$$\frac{d}{dt} \left(3 t^{2} - 7\right)$$$。
解答
和/差的导数等于导数的和/差：
$${\color{red}\left(\frac{d}{dt} \left(3 t^{2} - 7\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(3 t^{2}\right) - \frac{d}{dt} \left(7\right)\right)}$$对 $$$c = 3$$$ 和 $$$f{\left(t \right)} = t^{2}$$$ 应用常数倍法则 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dt} \left(3 t^{2}\right)\right)} - \frac{d}{dt} \left(7\right) = {\color{red}\left(3 \frac{d}{dt} \left(t^{2}\right)\right)} - \frac{d}{dt} \left(7\right)$$应用幂次法则 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，其中 $$$n = 2$$$:
$$3 {\color{red}\left(\frac{d}{dt} \left(t^{2}\right)\right)} - \frac{d}{dt} \left(7\right) = 3 {\color{red}\left(2 t\right)} - \frac{d}{dt} \left(7\right)$$常数的导数是$$$0$$$:
$$6 t - {\color{red}\left(\frac{d}{dt} \left(7\right)\right)} = 6 t - {\color{red}\left(0\right)}$$因此，$$$\frac{d}{dt} \left(3 t^{2} - 7\right) = 6 t$$$。
答案
$$$\frac{d}{dt} \left(3 t^{2} - 7\right) = 6 t$$$A
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