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$$$256 x^{2} + 16$$$的导数
您的输入
求$$$\frac{d}{dx} \left(256 x^{2} + 16\right)$$$。
解答
和/差的导数等于导数的和/差：
$${\color{red}\left(\frac{d}{dx} \left(256 x^{2} + 16\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(256 x^{2}\right) + \frac{d}{dx} \left(16\right)\right)}$$对 $$$c = 256$$$ 和 $$$f{\left(x \right)} = x^{2}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(256 x^{2}\right)\right)} + \frac{d}{dx} \left(16\right) = {\color{red}\left(256 \frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(16\right)$$常数的导数是$$$0$$$:
$${\color{red}\left(\frac{d}{dx} \left(16\right)\right)} + 256 \frac{d}{dx} \left(x^{2}\right) = {\color{red}\left(0\right)} + 256 \frac{d}{dx} \left(x^{2}\right)$$应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = 2$$$:
$$256 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = 256 {\color{red}\left(2 x\right)}$$因此，$$$\frac{d}{dx} \left(256 x^{2} + 16\right) = 512 x$$$。
答案
$$$\frac{d}{dx} \left(256 x^{2} + 16\right) = 512 x$$$A
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