$$$- \frac{2 x}{x^{2} + 1}$$$의 도함수
사용자 입력
$$$\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right)$$$을(를) 구하시오.
풀이
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:
$${\color{red}\left(\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right)\right)} = {\color{red}\left(- 2 \frac{d}{dx} \left(\frac{x}{x^{2} + 1}\right)\right)}$$$$$f{\left(x \right)} = x$$$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:
$$- 2 {\color{red}\left(\frac{d}{dx} \left(\frac{x}{x^{2} + 1}\right)\right)} = - 2 {\color{red}\left(\frac{\frac{d}{dx} \left(x\right) \left(x^{2} + 1\right) - x \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}\right)}$$멱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 1$$$에 대해 적용하면, 즉 $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$- \frac{2 \left(- x \frac{d}{dx} \left(x^{2} + 1\right) + \left(x^{2} + 1\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(- x \frac{d}{dx} \left(x^{2} + 1\right) + \left(x^{2} + 1\right) {\color{red}\left(1\right)}\right)}{\left(x^{2} + 1\right)^{2}}$$합/차의 도함수는 도함수들의 합/차이다:
$$- \frac{2 \left(x^{2} - x {\color{red}\left(\frac{d}{dx} \left(x^{2} + 1\right)\right)} + 1\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(x^{2} - x {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(1\right)\right)} + 1\right)}{\left(x^{2} + 1\right)^{2}}$$상수의 도함수는 $$$0$$$입니다:
$$- \frac{2 \left(x^{2} - x \left({\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x^{2}\right)\right) + 1\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(x^{2} - x \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)\right) + 1\right)}{\left(x^{2} + 1\right)^{2}}$$거듭제곱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 2$$$에 적용합니다:
$$- \frac{2 \left(x^{2} - x {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + 1\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(x^{2} - x {\color{red}\left(2 x\right)} + 1\right)}{\left(x^{2} + 1\right)^{2}}$$간단히 하시오:
$$- \frac{2 \left(1 - x^{2}\right)}{\left(x^{2} + 1\right)^{2}} = \frac{2 \left(x^{2} - 1\right)}{\left(x^{2} + 1\right)^{2}}$$따라서, $$$\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right) = \frac{2 \left(x^{2} - 1\right)}{\left(x^{2} + 1\right)^{2}}$$$.
정답
$$$\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right) = \frac{2 \left(x^{2} - 1\right)}{\left(x^{2} + 1\right)^{2}}$$$A