$$$\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}$$$ 的導數
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您的輸入
求$$$\frac{d}{dx} \left(\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}\right)$$$。
解答
函數 $$$\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}$$$ 是兩個函數 $$$f{\left(u \right)} = \operatorname{asin}{\left(u \right)}$$$ 與 $$$g{\left(x \right)} = \frac{2 x}{x^{2} + 1}$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。
應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\operatorname{asin}{\left(u \right)}\right) \frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right)\right)}$$反正弦函數的導數為 $$$\frac{d}{du} \left(\operatorname{asin}{\left(u \right)}\right) = \frac{1}{\sqrt{1 - u^{2}}}$$$:
$${\color{red}\left(\frac{d}{du} \left(\operatorname{asin}{\left(u \right)}\right)\right)} \frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right) = {\color{red}\left(\frac{1}{\sqrt{1 - u^{2}}}\right)} \frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right)$$返回原變數:
$$\frac{\frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right)}{\sqrt{1 - {\color{red}\left(u\right)}^{2}}} = \frac{\frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right)}{\sqrt{1 - {\color{red}\left(\frac{2 x}{x^{2} + 1}\right)}^{2}}}$$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 $$$c = 2$$$ 與 $$$f{\left(x \right)} = \frac{x}{x^{2} + 1}$$$:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right)\right)}}{\sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{{\color{red}\left(2 \frac{d}{dx} \left(\frac{x}{x^{2} + 1}\right)\right)}}{\sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$以 $$$f{\left(x \right)} = x$$$ 與 $$$g{\left(x \right)} = x^{2} + 1$$$ 套用商法則 $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$:
$$\frac{2 {\color{red}\left(\frac{d}{dx} \left(\frac{x}{x^{2} + 1}\right)\right)}}{\sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{2 {\color{red}\left(\frac{\frac{d}{dx} \left(x\right) \left(x^{2} + 1\right) - x \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}\right)}}{\sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$和/差的導數等於導數的和/差:
$$\frac{2 \left(- x {\color{red}\left(\frac{d}{dx} \left(x^{2} + 1\right)\right)} + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{2 \left(- x {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(1\right)\right)} + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$常數的導數為$$$0$$$:
$$\frac{2 \left(- x \left({\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x^{2}\right)\right) + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{2 \left(- x \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)\right) + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,取 $$$n = 2$$$:
$$\frac{2 \left(- x {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{2 \left(- x {\color{red}\left(2 x\right)} + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,取 $$$n = 1$$$,也就是 $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$\frac{2 \left(- 2 x^{2} + \left(x^{2} + 1\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{2 \left(- 2 x^{2} + \left(x^{2} + 1\right) {\color{red}\left(1\right)}\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$化簡:
$$\frac{2 \left(1 - x^{2}\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = - \frac{2 \left(x - 1\right) \left(x + 1\right)}{\left(x^{2} + 1\right) \left|{x - 1}\right| \left|{x + 1}\right|}$$因此,$$$\frac{d}{dx} \left(\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}\right) = - \frac{2 \left(x - 1\right) \left(x + 1\right)}{\left(x^{2} + 1\right) \left|{x - 1}\right| \left|{x + 1}\right|}$$$。
答案
$$$\frac{d}{dx} \left(\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}\right) = - \frac{2 \left(x - 1\right) \left(x + 1\right)}{\left(x^{2} + 1\right) \left|{x - 1}\right| \left|{x + 1}\right|}$$$A