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注意-y=ln(√(1+x²;)-x),∴e^(-y)=√(1+x²;)-x∴-e^(-y)=x-√(1+x²;) 而原式兩邊取指數函數,又有:e^y=x+√(1+x²;) 兩式相加:e^y-e^(-y)=2x ∴反函數為y=(e^x-e^(-x))/2
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