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$$$\operatorname{atan}{\left(4 x \right)}$$$的导数
您的输入
求$$$\frac{d}{dx} \left(\operatorname{atan}{\left(4 x \right)}\right)$$$。
解答
函数$$$\operatorname{atan}{\left(4 x \right)}$$$是两个函数$$$f{\left(u \right)} = \operatorname{atan}{\left(u \right)}$$$和$$$g{\left(x \right)} = 4 x$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(\operatorname{atan}{\left(4 x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right) \frac{d}{dx} \left(4 x\right)\right)}$$反正切函数的导数为 $$$\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right) = \frac{1}{u^{2} + 1}$$$:
$${\color{red}\left(\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right)\right)} \frac{d}{dx} \left(4 x\right) = {\color{red}\left(\frac{1}{u^{2} + 1}\right)} \frac{d}{dx} \left(4 x\right)$$返回到原变量：
$$\frac{\frac{d}{dx} \left(4 x\right)}{{\color{red}\left(u\right)}^{2} + 1} = \frac{\frac{d}{dx} \left(4 x\right)}{{\color{red}\left(4 x\right)}^{2} + 1}$$对 $$$c = 4$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$$\frac{{\color{red}\left(\frac{d}{dx} \left(4 x\right)\right)}}{16 x^{2} + 1} = \frac{{\color{red}\left(4 \frac{d}{dx} \left(x\right)\right)}}{16 x^{2} + 1}$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：
$$\frac{4 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{16 x^{2} + 1} = \frac{4 {\color{red}\left(1\right)}}{16 x^{2} + 1}$$因此，$$$\frac{d}{dx} \left(\operatorname{atan}{\left(4 x \right)}\right) = \frac{4}{16 x^{2} + 1}$$$。
答案
$$$\frac{d}{dx} \left(\operatorname{atan}{\left(4 x \right)}\right) = \frac{4}{16 x^{2} + 1}$$$A