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$$$x^{2} - 1$$$的导数
您的输入
求$$$\frac{d}{dx} \left(x^{2} - 1\right)$$$。
解答
和/差的导数等于导数的和/差：
$${\color{red}\left(\frac{d}{dx} \left(x^{2} - 1\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(1\right)\right)}$$常数的导数是$$$0$$$:
$$- {\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x^{2}\right) = - {\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)$$应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = 2$$$:
$${\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = {\color{red}\left(2 x\right)}$$因此，$$$\frac{d}{dx} \left(x^{2} - 1\right) = 2 x$$$。
答案
$$$\frac{d}{dx} \left(x^{2} - 1\right) = 2 x$$$A
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