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