$$$\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}$$$의 도함수

함수 $$$\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}$$$는 두 함수 $$$f{\left(u \right)} = \operatorname{asin}{\left(u \right)}$$$와 $$$g{\left(x \right)} = \frac{2 x}{x^{2} + 1}$$$의 합성함수 $$$f{\left(g{\left(x \right)} \right)}$$$이다.

연쇄법칙 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$을(를) 적용하십시오:

$${\color{red}\left(\frac{d}{dx} \left(\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\operatorname{asin}{\left(u \right)}\right) \frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right)\right)}$$

아크사인 함수의 도함수는 $$$\frac{d}{du} \left(\operatorname{asin}{\left(u \right)}\right) = \frac{1}{\sqrt{1 - u^{2}}}$$$:

$${\color{red}\left(\frac{d}{du} \left(\operatorname{asin}{\left(u \right)}\right)\right)} \frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right) = {\color{red}\left(\frac{1}{\sqrt{1 - u^{2}}}\right)} \frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right)$$

역치환:

$$\frac{\frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right)}{\sqrt{1 - {\color{red}\left(u\right)}^{2}}} = \frac{\frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right)}{\sqrt{1 - {\color{red}\left(\frac{2 x}{x^{2} + 1}\right)}^{2}}}$$

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = 2$$$와 $$$f{\left(x \right)} = \frac{x}{x^{2} + 1}$$$에 적용합니다:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(\frac{2 x}{x^{2} + 1}\right)\right)}}{\sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{{\color{red}\left(2 \frac{d}{dx} \left(\frac{x}{x^{2} + 1}\right)\right)}}{\sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$

$$$f{\left(x \right)} = x$$$와 $$$g{\left(x \right)} = x^{2} + 1$$$에 대해 몫의 미분법칙 $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$을 적용하십시오:

$$\frac{2 {\color{red}\left(\frac{d}{dx} \left(\frac{x}{x^{2} + 1}\right)\right)}}{\sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{2 {\color{red}\left(\frac{\frac{d}{dx} \left(x\right) \left(x^{2} + 1\right) - x \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}\right)}}{\sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$

합/차의 도함수는 도함수들의 합/차이다:

$$\frac{2 \left(- x {\color{red}\left(\frac{d}{dx} \left(x^{2} + 1\right)\right)} + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{2 \left(- x {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(1\right)\right)} + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$

상수의 도함수는 $$$0$$$입니다:

$$\frac{2 \left(- x \left({\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x^{2}\right)\right) + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{2 \left(- x \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)\right) + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$

거듭제곱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 2$$$에 적용합니다:

$$\frac{2 \left(- x {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{2 \left(- x {\color{red}\left(2 x\right)} + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$

멱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 1$$$에 대해 적용하면, 즉 $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$\frac{2 \left(- 2 x^{2} + \left(x^{2} + 1\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = \frac{2 \left(- 2 x^{2} + \left(x^{2} + 1\right) {\color{red}\left(1\right)}\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}}$$

간단히 하시오:

$$\frac{2 \left(1 - x^{2}\right)}{\left(x^{2} + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x^{2} + 1\right)^{2}} + 1}} = - \frac{2 \left(x - 1\right) \left(x + 1\right)}{\left(x^{2} + 1\right) \left|{x - 1}\right| \left|{x + 1}\right|}$$

따라서, $$$\frac{d}{dx} \left(\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}\right) = - \frac{2 \left(x - 1\right) \left(x + 1\right)}{\left(x^{2} + 1\right) \left|{x - 1}\right| \left|{x + 1}\right|}.$$$