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$$$x^{2} + 1$$$ 的導數
您的輸入
求$$$\frac{d}{dx} \left(x^{2} + 1\right)$$$。
解答
和/差的導數等於導數的和/差：
$${\color{red}\left(\frac{d}{dx} \left(x^{2} + 1\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(1\right)\right)}$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 2$$$：
$${\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(1\right) = {\color{red}\left(2 x\right)} + \frac{d}{dx} \left(1\right)$$常數的導數為$$$0$$$：
$$2 x + {\color{red}\left(\frac{d}{dx} \left(1\right)\right)} = 2 x + {\color{red}\left(0\right)}$$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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