$$$\sqrt{2} u - 1$$$的导数
您的输入
求$$$\frac{d}{du} \left(\sqrt{2} u - 1\right)$$$。
解答
和/差的导数等于导数的和/差：
$${\color{red}\left(\frac{d}{du} \left(\sqrt{2} u - 1\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\sqrt{2} u\right) - \frac{d}{du} \left(1\right)\right)}$$常数的导数是$$$0$$$:
$$- {\color{red}\left(\frac{d}{du} \left(1\right)\right)} + \frac{d}{du} \left(\sqrt{2} u\right) = - {\color{red}\left(0\right)} + \frac{d}{du} \left(\sqrt{2} u\right)$$对 $$$c = \sqrt{2}$$$ 和 $$$f{\left(u \right)} = u$$$ 应用常数倍法则 $$$\frac{d}{du} \left(c f{\left(u \right)}\right) = c \frac{d}{du} \left(f{\left(u \right)}\right)$$$：
$${\color{red}\left(\frac{d}{du} \left(\sqrt{2} u\right)\right)} = {\color{red}\left(\sqrt{2} \frac{d}{du} \left(u\right)\right)}$$应用幂法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{du} \left(u\right) = 1$$$：
$$\sqrt{2} {\color{red}\left(\frac{d}{du} \left(u\right)\right)} = \sqrt{2} {\color{red}\left(1\right)}$$因此，$$$\frac{d}{du} \left(\sqrt{2} u - 1\right) = \sqrt{2}$$$。
答案
$$$\frac{d}{du} \left(\sqrt{2} u - 1\right) = \sqrt{2}$$$A
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