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$$$\sqrt{x}$$$ 的二階導數
您的輸入
求$$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right)$$$。
解答
求第一階導數 $$$\frac{d}{dx} \left(\sqrt{x}\right)$$$
套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = \frac{1}{2}$$$：
$${\color{red}\left(\frac{d}{dx} \left(\sqrt{x}\right)\right)} = {\color{red}\left(\frac{1}{2 \sqrt{x}}\right)}$$因此，$$$\frac{d}{dx} \left(\sqrt{x}\right) = \frac{1}{2 \sqrt{x}}$$$。
接下來，$$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right) = \frac{d}{dx} \left(\frac{1}{2 \sqrt{x}}\right)$$$
套用常數倍法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$，使用 $$$c = \frac{1}{2}$$$ 與 $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$：
$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{2 \sqrt{x}}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right)}{2}\right)}$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = - \frac{1}{2}$$$：
$$\frac{{\color{red}\left(\frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right)\right)}}{2} = \frac{{\color{red}\left(- \frac{1}{2 x^{\frac{3}{2}}}\right)}}{2}$$因此，$$$\frac{d}{dx} \left(\frac{1}{2 \sqrt{x}}\right) = - \frac{1}{4 x^{\frac{3}{2}}}$$$。
因此，$$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right) = - \frac{1}{4 x^{\frac{3}{2}}}$$$。
答案
$$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right) = - \frac{1}{4 x^{\frac{3}{2}}}$$$A
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