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March 20, 2026 01:33
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여러 가지 미분 공식 Anki Deck
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| \(y = c\) \(y' = 0\)<br><br>조건: \(x∈ℝ\)<br><br>증명: \(\frac{c-c}{h}=0\) | |
| \(y = x^n\) \(y' = nx^{n-1}\)<br><br>조건: \(n∈ℝ\) (일반적으로 \(x>0\))<br><br>증명: 이항정리 | |
| \(y = f(x)g(x)\) \(y' = f'g + fg'\)<br><br>조건: 미분가능<br><br>증명: 정의 전개 | |
| \(y = \frac{g(x)}{f(x)}\) \(y' = \frac{g'f - gf'}{f^2}\)<br><br>조건: \(f≠0\)<br><br>증명: 곱미분 활용 | |
| \(y = \sin x\) \(y' = \cos x\)<br><br>조건: \(ℝ\)<br><br>증명: \(\lim \frac{\sin h}{h}=1\) | |
| \(y = \cos x\) \(y' = -\sin x\)<br><br>조건: \(ℝ\)<br><br>증명: 삼각항등식 | |
| \(y = \tan x\) \(y' = \sec^2 x\)<br><br>조건: \(x≠\frac{\pi}{2}+k\pi\)<br><br>증명: 몫미분 | |
| \(y = \cot x\) \(y' = -\csc^2 x\)<br><br>조건: \(x≠k\pi\)<br><br>증명: 몫미분 | |
| \(y = \sec x\) \(y' = \sec x\tan x\)<br><br>조건: \(\cos x≠0\)<br><br>증명: 역수미분 | |
| \(y = \csc x\) \(y' = -\csc x\cot x\)<br><br>조건: \(\sin x≠0\)<br><br>증명: 역수미분 | |
| \(y = \sin^{-1} x\) \(y' = \frac{1}{\sqrt{1-x^2}}\)<br><br>조건: \(|x|<1\)<br><br>증명: \(x=\sin y\) | |
| \(y = \cos^{-1} x\) \(y' = -\frac{1}{\sqrt{1-x^2}}\)<br><br>조건: \(|x|<1\)<br><br>증명: \(x=\cos y\) | |
| \(y = \tan^{-1} x\) \(y' = \frac{1}{1+x^2}\)<br><br>조건: \(ℝ\)<br><br>증명: \(x=\tan y\) | |
| \(y = \cot^{-1} x\) \(y' = -\frac{1}{1+x^2}\)<br><br>조건: \(ℝ\)<br><br>증명: \(x=\cot y\) | |
| \(y = \sec^{-1} x\) \(y' = \frac{1}{|x|\sqrt{x^2-1}}\)<br><br>조건: \(|x|>1\)<br><br>증명: \(x=\sec y\) | |
| \(y = \csc^{-1} x\) \(y' = -\frac{1}{|x|\sqrt{x^2-1}}\)<br><br>조건: \(|x|>1\)<br><br>증명: \(x=\csc y\) | |
| \(y = a^x\) \(y' = a^x\ln a\)<br><br>조건: \(a>0,a≠1\)<br><br>증명: \(e^{x\ln a}\) | |
| \(y = \log_a x\) \(y' = \frac{1}{x\ln a}\)<br><br>조건: \(x>0\)<br><br>증명: ln 변환 | |
| \(y = e^x\) \(y' = e^x\)<br><br>조건: \(ℝ\)<br><br>증명: 정의 | |
| \(y = \ln x\) \(y' = \frac{1}{x}\)<br><br>조건: \(x>0\)<br><br>증명: 역함수 미분 | |
| \(y = \sinh x\) \(y' = \cosh x\)<br><br>조건: \(ℝ\)<br><br>증명: 정의 | |
| \(y = \cosh x\) \(y' = \sinh x\)<br><br>조건: \(ℝ\)<br><br>증명: 정의 | |
| \(y = \tanh x\) \(y' = \operatorname{sech}^2 x\)<br><br>조건: \(ℝ\)<br><br>증명: 몫미분 | |
| \(y = \operatorname{sech} x\) \(y' = -\operatorname{sech} x\tanh x\)<br><br>조건: \(ℝ\)<br><br>증명: 역수 | |
| \(y = \operatorname{csch} x\) \(y' = -\operatorname{csch} x\coth x\)<br><br>조건: \(x≠0\)<br><br>증명: 역수 | |
| \(y = \coth x\) \(y' = -\operatorname{csch}^2 x\)<br><br>조건: \(x≠0\)<br><br>증명: 몫미분 | |
| \(y = \sinh^{-1} x\) \(y' = \frac{1}{\sqrt{1+x^2}}\)<br><br>조건: \(ℝ\)<br><br>증명: \(x=\sinh y\) | |
| \(y = \cosh^{-1} x\) \(y' = \frac{1}{\sqrt{x^2-1}}\)<br><br>조건: \(x>1\)<br><br>증명: \(x=\cosh y\) | |
| \(y = \tanh^{-1} x\) \(y' = \frac{1}{1-x^2}\)<br><br>조건: \(|x|<1\)<br><br>증명: \(x=\tanh y\) | |
| \(y = \coth^{-1} x\) \(y' = \frac{1}{1-x^2}\)<br><br>조건: \(|x|>1\)<br><br>증명: \(x=\coth y\) | |
| \(y = \operatorname{sech}^{-1} x\) \(y' = -\frac{1}{x\sqrt{1-x^2}}\)<br><br>조건: \(0<x<1\)<br><br>증명: \(x=\sech y\) | |
| \(y = \operatorname{csch}^{-1} x\) \(y' = -\frac{1}{|x|\sqrt{1+x^2}}\)<br><br>조건: \(x≠0\)<br><br>증명: \(x=\csch y\) |
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