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$$$9 t^{2} + 4$$$の導関数
入力内容
$$$\frac{d}{dt} \left(9 t^{2} + 4\right)$$$ を求めよ。
解答
和/差の導関数は、導関数の和/差である:
$${\color{red}\left(\frac{d}{dt} \left(9 t^{2} + 4\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(9 t^{2}\right) + \frac{d}{dt} \left(4\right)\right)}$$定数の導数は$$$0$$$です:
$${\color{red}\left(\frac{d}{dt} \left(4\right)\right)} + \frac{d}{dt} \left(9 t^{2}\right) = {\color{red}\left(0\right)} + \frac{d}{dt} \left(9 t^{2}\right)$$定数倍の法則 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ を $$$c = 9$$$ と $$$f{\left(t \right)} = t^{2}$$$ に対して適用します:
$${\color{red}\left(\frac{d}{dt} \left(9 t^{2}\right)\right)} = {\color{red}\left(9 \frac{d}{dt} \left(t^{2}\right)\right)}$$冪法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ を $$$n = 2$$$ に対して適用する:
$$9 {\color{red}\left(\frac{d}{dt} \left(t^{2}\right)\right)} = 9 {\color{red}\left(2 t\right)}$$したがって、$$$\frac{d}{dt} \left(9 t^{2} + 4\right) = 18 t$$$。
解答
$$$\frac{d}{dt} \left(9 t^{2} + 4\right) = 18 t$$$A
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