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$$$x^{3} - 2 x$$$ 在 $$$x = c$$$ 处的导数
您的输入
求出$$$\frac{d}{dx} \left(x^{3} - 2 x\right)$$$，并在$$$x = c$$$处计算其值。
解答
和/差的导数等于导数的和/差：
$${\color{red}\left(\frac{d}{dx} \left(x^{3} - 2 x\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) - \frac{d}{dx} \left(2 x\right)\right)}$$对 $$$c = 2$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$$- {\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} + \frac{d}{dx} \left(x^{3}\right) = - {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(x^{3}\right)$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：
$$- 2 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(x^{3}\right) = - 2 {\color{red}\left(1\right)} + \frac{d}{dx} \left(x^{3}\right)$$应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = 3$$$:
$${\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)} - 2 = {\color{red}\left(3 x^{2}\right)} - 2$$因此，$$$\frac{d}{dx} \left(x^{3} - 2 x\right) = 3 x^{2} - 2$$$。
最后，在$$$x = c$$$处计算导数的值。
$$$\left(\frac{d}{dx} \left(x^{3} - 2 x\right)\right)|_{\left(x = c\right)} = 3 c^{2} - 2$$$
答案
$$$\frac{d}{dx} \left(x^{3} - 2 x\right) = 3 x^{2} - 2$$$A
$$$\left(\frac{d}{dx} \left(x^{3} - 2 x\right)\right)|_{\left(x = c\right)} = 3 c^{2} - 2$$$A