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$$$\frac{1}{x}$$$ 的二阶导数
您的输入
求$$$\frac{d^{2}}{dx^{2}} \left(\frac{1}{x}\right)$$$。
解答
求一阶导数 $$$\frac{d}{dx} \left(\frac{1}{x}\right)$$$
应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = -1$$$:
$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{x}\right)\right)} = {\color{red}\left(- \frac{1}{x^{2}}\right)}$$因此，$$$\frac{d}{dx} \left(\frac{1}{x}\right) = - \frac{1}{x^{2}}$$$。
接下来，$$$\frac{d^{2}}{dx^{2}} \left(\frac{1}{x}\right) = \frac{d}{dx} \left(- \frac{1}{x^{2}}\right)$$$
对 $$$c = -1$$$ 和 $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(- \frac{1}{x^{2}}\right)\right)} = {\color{red}\left(- \frac{d}{dx} \left(\frac{1}{x^{2}}\right)\right)}$$应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = -2$$$:
$$- {\color{red}\left(\frac{d}{dx} \left(\frac{1}{x^{2}}\right)\right)} = - {\color{red}\left(- \frac{2}{x^{3}}\right)}$$因此，$$$\frac{d}{dx} \left(- \frac{1}{x^{2}}\right) = \frac{2}{x^{3}}$$$。
因此，$$$\frac{d^{2}}{dx^{2}} \left(\frac{1}{x}\right) = \frac{2}{x^{3}}$$$。
答案
$$$\frac{d^{2}}{dx^{2}} \left(\frac{1}{x}\right) = \frac{2}{x^{3}}$$$A
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