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$$$t + \sqrt{2}$$$の導関数
入力内容
$$$\frac{d}{dt} \left(t + \sqrt{2}\right)$$$ を求めよ。
解答
和/差の導関数は、導関数の和/差である:
$${\color{red}\left(\frac{d}{dt} \left(t + \sqrt{2}\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(t\right) + \frac{d}{dt} \left(\sqrt{2}\right)\right)}$$定数の導数は$$$0$$$です:
$${\color{red}\left(\frac{d}{dt} \left(\sqrt{2}\right)\right)} + \frac{d}{dt} \left(t\right) = {\color{red}\left(0\right)} + \frac{d}{dt} \left(t\right)$$$$$n = 1$$$ を用いて冪法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ を適用すると、すなわち $$$\frac{d}{dt} \left(t\right) = 1$$$:
$${\color{red}\left(\frac{d}{dt} \left(t\right)\right)} = {\color{red}\left(1\right)}$$したがって、$$$\frac{d}{dt} \left(t + \sqrt{2}\right) = 1$$$。
解答
$$$\frac{d}{dt} \left(t + \sqrt{2}\right) = 1$$$A
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