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$$$9 x e^{2} - 4$$$的导数
您的输入
求$$$\frac{d}{dx} \left(9 x e^{2} - 4\right)$$$。
解答
和/差的导数等于导数的和/差：
$${\color{red}\left(\frac{d}{dx} \left(9 x e^{2} - 4\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(9 x e^{2}\right) - \frac{d}{dx} \left(4\right)\right)}$$对 $$$c = 9 e^{2}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(9 x e^{2}\right)\right)} - \frac{d}{dx} \left(4\right) = {\color{red}\left(9 e^{2} \frac{d}{dx} \left(x\right)\right)} - \frac{d}{dx} \left(4\right)$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：
$$9 e^{2} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} - \frac{d}{dx} \left(4\right) = 9 e^{2} {\color{red}\left(1\right)} - \frac{d}{dx} \left(4\right)$$常数的导数是$$$0$$$:
$$- {\color{red}\left(\frac{d}{dx} \left(4\right)\right)} + 9 e^{2} = - {\color{red}\left(0\right)} + 9 e^{2}$$因此，$$$\frac{d}{dx} \left(9 x e^{2} - 4\right) = 9 e^{2}$$$。
答案
$$$\frac{d}{dx} \left(9 x e^{2} - 4\right) = 9 e^{2}$$$A
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