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$$$e^{- \frac{1}{u^{2}}}$$$의 도함수
사용자 입력
$$$\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right)$$$을(를) 구하시오.
풀이
함수 $$$e^{- \frac{1}{u^{2}}}$$$는 두 함수 $$$f{\left(v \right)} = e^{v}$$$와 $$$g{\left(u \right)} = - \frac{1}{u^{2}}$$$의 합성함수 $$$f{\left(g{\left(u \right)} \right)}$$$이다.
연쇄법칙 $$$\frac{d}{du} \left(f{\left(g{\left(u \right)} \right)}\right) = \frac{d}{dv} \left(f{\left(v \right)}\right) \frac{d}{du} \left(g{\left(u \right)}\right)$$$을(를) 적용하십시오:
$${\color{red}\left(\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right)\right)} = {\color{red}\left(\frac{d}{dv} \left(e^{v}\right) \frac{d}{du} \left(- \frac{1}{u^{2}}\right)\right)}$$지수함수의 도함수는 $$$\frac{d}{dv} \left(e^{v}\right) = e^{v}$$$:
$${\color{red}\left(\frac{d}{dv} \left(e^{v}\right)\right)} \frac{d}{du} \left(- \frac{1}{u^{2}}\right) = {\color{red}\left(e^{v}\right)} \frac{d}{du} \left(- \frac{1}{u^{2}}\right)$$역치환:
$$e^{{\color{red}\left(v\right)}} \frac{d}{du} \left(- \frac{1}{u^{2}}\right) = e^{{\color{red}\left(- \frac{1}{u^{2}}\right)}} \frac{d}{du} \left(- \frac{1}{u^{2}}\right)$$상수배 법칙 $$$\frac{d}{du} \left(c f{\left(u \right)}\right) = c \frac{d}{du} \left(f{\left(u \right)}\right)$$$을 $$$c = -1$$$와 $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$에 적용합니다:
$$e^{- \frac{1}{u^{2}}} {\color{red}\left(\frac{d}{du} \left(- \frac{1}{u^{2}}\right)\right)} = e^{- \frac{1}{u^{2}}} {\color{red}\left(- \frac{d}{du} \left(\frac{1}{u^{2}}\right)\right)}$$거듭제곱법칙 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$을 $$$n = -2$$$에 적용합니다:
$$- e^{- \frac{1}{u^{2}}} {\color{red}\left(\frac{d}{du} \left(\frac{1}{u^{2}}\right)\right)} = - e^{- \frac{1}{u^{2}}} {\color{red}\left(- \frac{2}{u^{3}}\right)}$$따라서, $$$\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right) = \frac{2 e^{- \frac{1}{u^{2}}}}{u^{3}}$$$.
정답
$$$\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right) = \frac{2 e^{- \frac{1}{u^{2}}}}{u^{3}}$$$A