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$$$9 t^{2} + 4$$$ 的導數
您的輸入
求$$$\frac{d}{dt} \left(9 t^{2} + 4\right)$$$。
解答
和/差的導數等於導數的和/差：
$${\color{red}\left(\frac{d}{dt} \left(9 t^{2} + 4\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(9 t^{2}\right) + \frac{d}{dt} \left(4\right)\right)}$$常數的導數為$$$0$$$：
$${\color{red}\left(\frac{d}{dt} \left(4\right)\right)} + \frac{d}{dt} \left(9 t^{2}\right) = {\color{red}\left(0\right)} + \frac{d}{dt} \left(9 t^{2}\right)$$套用常數倍法則 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$，使用 $$$c = 9$$$ 與 $$$f{\left(t \right)} = t^{2}$$$：
$${\color{red}\left(\frac{d}{dt} \left(9 t^{2}\right)\right)} = {\color{red}\left(9 \frac{d}{dt} \left(t^{2}\right)\right)}$$套用冪次法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，取 $$$n = 2$$$：
$$9 {\color{red}\left(\frac{d}{dt} \left(t^{2}\right)\right)} = 9 {\color{red}\left(2 t\right)}$$因此，$$$\frac{d}{dt} \left(9 t^{2} + 4\right) = 18 t$$$。
答案
$$$\frac{d}{dt} \left(9 t^{2} + 4\right) = 18 t$$$A
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