$$$\frac{1}{1 + e^{- x}}$$$의 이차 도함수

제1도함수 $$$\frac{d}{dx} \left(\frac{1}{1 + e^{- x}}\right)$$$를 구하세요

함수 $$$\frac{1}{1 + e^{- x}}$$$는 두 함수 $$$f{\left(u \right)} = \frac{1}{u}$$$와 $$$g{\left(x \right)} = 1 + e^{- x}$$$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:

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

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

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

역치환:

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

합/차의 도함수는 도함수들의 합/차이다:

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

상수의 도함수는 $$$0$$$입니다:

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

함수 $$$e^{- x}$$$는 두 함수 $$$f{\left(u \right)} = e^{u}$$$와 $$$g{\left(x \right)} = - x$$$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:

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

지수함수의 도함수는 $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:

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

역치환:

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

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

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

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

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

간단히 하시오:

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

따라서, $$$\frac{d}{dx} \left(\frac{1}{1 + e^{- x}}\right) = \frac{1}{4 \cosh^{2}{\left(\frac{x}{2} \right)}}$$$.

다음으로, $$$\frac{d^{2}}{dx^{2}} \left(\frac{1}{1 + e^{- x}}\right) = \frac{d}{dx} \left(\frac{1}{4 \cosh^{2}{\left(\frac{x}{2} \right)}}\right)$$$

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = \frac{1}{4}$$$와 $$$f{\left(x \right)} = \frac{1}{\cosh^{2}{\left(\frac{x}{2} \right)}}$$$에 적용합니다:

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

함수 $$$\frac{1}{\cosh^{2}{\left(\frac{x}{2} \right)}}$$$는 두 함수 $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$와 $$$g{\left(x \right)} = \cosh{\left(\frac{x}{2} \right)}$$$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:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(\frac{1}{\cosh^{2}{\left(\frac{x}{2} \right)}}\right)\right)}}{4} = \frac{{\color{red}\left(\frac{d}{du} \left(\frac{1}{u^{2}}\right) \frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)\right)}}{4}$$

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

$$\frac{{\color{red}\left(\frac{d}{du} \left(\frac{1}{u^{2}}\right)\right)} \frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)}{4} = \frac{{\color{red}\left(- \frac{2}{u^{3}}\right)} \frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)}{4}$$

역치환:

$$- \frac{\frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)}{2 {\color{red}\left(u\right)}^{3}} = - \frac{\frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)}{2 {\color{red}\left(\cosh{\left(\frac{x}{2} \right)}\right)}^{3}}$$

함수 $$$\cosh{\left(\frac{x}{2} \right)}$$$는 두 함수 $$$f{\left(u \right)} = \cosh{\left(u \right)}$$$와 $$$g{\left(x \right)} = \frac{x}{2}$$$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:

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)\right)}}{2 \cosh^{3}{\left(\frac{x}{2} \right)}} = - \frac{{\color{red}\left(\frac{d}{du} \left(\cosh{\left(u \right)}\right) \frac{d}{dx} \left(\frac{x}{2}\right)\right)}}{2 \cosh^{3}{\left(\frac{x}{2} \right)}}$$

쌍곡선 코사인 함수의 도함수는 $$$\frac{d}{du} \left(\cosh{\left(u \right)}\right) = \sinh{\left(u \right)}$$$:

$$- \frac{{\color{red}\left(\frac{d}{du} \left(\cosh{\left(u \right)}\right)\right)} \frac{d}{dx} \left(\frac{x}{2}\right)}{2 \cosh^{3}{\left(\frac{x}{2} \right)}} = - \frac{{\color{red}\left(\sinh{\left(u \right)}\right)} \frac{d}{dx} \left(\frac{x}{2}\right)}{2 \cosh^{3}{\left(\frac{x}{2} \right)}}$$

역치환:

$$- \frac{\sinh{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(\frac{x}{2}\right)}{2 \cosh^{3}{\left(\frac{x}{2} \right)}} = - \frac{\sinh{\left({\color{red}\left(\frac{x}{2}\right)} \right)} \frac{d}{dx} \left(\frac{x}{2}\right)}{2 \cosh^{3}{\left(\frac{x}{2} \right)}}$$

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

$$- \frac{\sinh{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{d}{dx} \left(\frac{x}{2}\right)\right)}}{2 \cosh^{3}{\left(\frac{x}{2} \right)}} = - \frac{\sinh{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{\frac{d}{dx} \left(x\right)}{2}\right)}}{2 \cosh^{3}{\left(\frac{x}{2} \right)}}$$

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

$$- \frac{\sinh{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{4 \cosh^{3}{\left(\frac{x}{2} \right)}} = - \frac{\sinh{\left(\frac{x}{2} \right)} {\color{red}\left(1\right)}}{4 \cosh^{3}{\left(\frac{x}{2} \right)}}$$

따라서, $$$\frac{d}{dx} \left(\frac{1}{4 \cosh^{2}{\left(\frac{x}{2} \right)}}\right) = - \frac{\sinh{\left(\frac{x}{2} \right)}}{4 \cosh^{3}{\left(\frac{x}{2} \right)}}$$$.

따라서 $$$\frac{d^{2}}{dx^{2}} \left(\frac{1}{1 + e^{- x}}\right) = - \frac{\sinh{\left(\frac{x}{2} \right)}}{4 \cosh^{3}{\left(\frac{x}{2} \right)}}$$$.