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$$$256 x^{2} + 16$$$ 的導數
您的輸入
求$$$\frac{d}{dx} \left(256 x^{2} + 16\right)$$$。
解答
和/差的導數等於導數的和/差：
$${\color{red}\left(\frac{d}{dx} \left(256 x^{2} + 16\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(256 x^{2}\right) + \frac{d}{dx} \left(16\right)\right)}$$套用常數倍法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$，使用 $$$c = 256$$$ 與 $$$f{\left(x \right)} = x^{2}$$$：
$${\color{red}\left(\frac{d}{dx} \left(256 x^{2}\right)\right)} + \frac{d}{dx} \left(16\right) = {\color{red}\left(256 \frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(16\right)$$常數的導數為$$$0$$$：
$${\color{red}\left(\frac{d}{dx} \left(16\right)\right)} + 256 \frac{d}{dx} \left(x^{2}\right) = {\color{red}\left(0\right)} + 256 \frac{d}{dx} \left(x^{2}\right)$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 2$$$：
$$256 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = 256 {\color{red}\left(2 x\right)}$$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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