$$$4 m^{2} + 16 x^{2}$$$ 关于 $$$x$$$ 的导数
您的输入
求$$$\frac{d}{dx} \left(4 m^{2} + 16 x^{2}\right)$$$。
解答
和/差的导数等于导数的和/差：
$${\color{red}\left(\frac{d}{dx} \left(4 m^{2} + 16 x^{2}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(4 m^{2}\right) + \frac{d}{dx} \left(16 x^{2}\right)\right)}$$常数的导数是$$$0$$$:
$${\color{red}\left(\frac{d}{dx} \left(4 m^{2}\right)\right)} + \frac{d}{dx} \left(16 x^{2}\right) = {\color{red}\left(0\right)} + \frac{d}{dx} \left(16 x^{2}\right)$$对 $$$c = 16$$$ 和 $$$f{\left(x \right)} = x^{2}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(16 x^{2}\right)\right)} = {\color{red}\left(16 \frac{d}{dx} \left(x^{2}\right)\right)}$$应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = 2$$$:
$$16 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = 16 {\color{red}\left(2 x\right)}$$因此，$$$\frac{d}{dx} \left(4 m^{2} + 16 x^{2}\right) = 32 x$$$。
答案
$$$\frac{d}{dx} \left(4 m^{2} + 16 x^{2}\right) = 32 x$$$A