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函數 $$$x^{3} - 2 x$$$ 在 $$$x = c$$$ 處的導數
您的輸入
求出 $$$\frac{d}{dx} \left(x^{3} - 2 x\right)$$$，並在 $$$x = c$$$ 處求值。
解答
和/差的導數等於導數的和/差：
$${\color{red}\left(\frac{d}{dx} \left(x^{3} - 2 x\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) - \frac{d}{dx} \left(2 x\right)\right)}$$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 $$$c = 2$$$ 與 $$$f{\left(x \right)} = x$$$：
$$- {\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} + \frac{d}{dx} \left(x^{3}\right) = - {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(x^{3}\right)$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是 $$$\frac{d}{dx} \left(x\right) = 1$$$：
$$- 2 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(x^{3}\right) = - 2 {\color{red}\left(1\right)} + \frac{d}{dx} \left(x^{3}\right)$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 3$$$：
$${\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)} - 2 = {\color{red}\left(3 x^{2}\right)} - 2$$因此，$$$\frac{d}{dx} \left(x^{3} - 2 x\right) = 3 x^{2} - 2$$$。
最後，在$$$x = c$$$處計算導數。
$$$\left(\frac{d}{dx} \left(x^{3} - 2 x\right)\right)|_{\left(x = c\right)} = 3 c^{2} - 2$$$
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