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