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$$$\operatorname{atan}{\left(4 x \right)}$$$ 的導數
您的輸入
求$$$\frac{d}{dx} \left(\operatorname{atan}{\left(4 x \right)}\right)$$$。
解答
函數 $$$\operatorname{atan}{\left(4 x \right)}$$$ 是兩個函數 $$$f{\left(u \right)} = \operatorname{atan}{\left(u \right)}$$$ 與 $$$g{\left(x \right)} = 4 x$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。
應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(\operatorname{atan}{\left(4 x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right) \frac{d}{dx} \left(4 x\right)\right)}$$反正切函數的導數是 $$$\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right) = \frac{1}{u^{2} + 1}$$$:
$${\color{red}\left(\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right)\right)} \frac{d}{dx} \left(4 x\right) = {\color{red}\left(\frac{1}{u^{2} + 1}\right)} \frac{d}{dx} \left(4 x\right)$$返回原變數：
$$\frac{\frac{d}{dx} \left(4 x\right)}{{\color{red}\left(u\right)}^{2} + 1} = \frac{\frac{d}{dx} \left(4 x\right)}{{\color{red}\left(4 x\right)}^{2} + 1}$$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 $$$c = 4$$$ 與 $$$f{\left(x \right)} = x$$$：
$$\frac{{\color{red}\left(\frac{d}{dx} \left(4 x\right)\right)}}{16 x^{2} + 1} = \frac{{\color{red}\left(4 \frac{d}{dx} \left(x\right)\right)}}{16 x^{2} + 1}$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是 $$$\frac{d}{dx} \left(x\right) = 1$$$：
$$\frac{4 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{16 x^{2} + 1} = \frac{4 {\color{red}\left(1\right)}}{16 x^{2} + 1}$$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