|
|
|
|
|
|
|
|
|
|
|
|
$$$x^{2} + 1$$$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：
$${\color{red}\left(\frac{d}{dx} \left(x^{2} + 1\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(1\right)\right)}$$应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = 2$$$:
$${\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(1\right) = {\color{red}\left(2 x\right)} + \frac{d}{dx} \left(1\right)$$常数的导数是$$$0$$$:
$$2 x + {\color{red}\left(\frac{d}{dx} \left(1\right)\right)} = 2 x + {\color{red}\left(0\right)}$$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
Please try a new game Rotatly