反雙曲函數是雙曲函數的反函數。與反圓函數不同之處是它的前綴是ar意即area(面積),而不是arc(弧)。因為雙曲角是以雙曲線、通過原點直線以及其對x軸的映射三者之間所夾面積定義的,而圓角是以弧長與半徑的比值定義。 反雙曲函數示意圖 由於已知的技術原因,圖表暫時不可用。帶來不便,我們深表歉意。 幾個反雙曲函數的圖形 符號 s i n h − 1 , c o s h − 1 {\displaystyle \mathrm {sinh} ^{-1},\mathrm {cosh} ^{-1}} 等常用於 a r s i n h , a r c o s h {\displaystyle \mathrm {arsinh} ,\mathrm {arcosh} } 等。但是這種符號有時在 s i n h − 1 x {\displaystyle \mathrm {sinh} ^{-1}x} 和 1 s i n h x {\displaystyle {\frac {1}{\mathrm {sinh} x}}} 之間易造成混淆。 Remove ads 下表列出基本的反雙曲函數。 更多資訊 , ... 名稱 常用符號 定義 定義域 值域 圖像 反雙曲正弦 y = a r s i n h x {\displaystyle y=\mathrm {arsinh} x} ln ( x + x 2 + 1 ) {\displaystyle \ln(x+{\sqrt {x^{2}+1}})} R {\displaystyle \mathbb {R} } R {\displaystyle \mathbb {R} } 反雙曲餘弦 y = a r c o s h x {\displaystyle y=\mathrm {arcosh} x} ln ( x ± x 2 − 1 ) {\displaystyle \ln(x\pm {\sqrt {x^{2}-1}})} [註 1] [ 1 , + ∞ ) {\displaystyle [1,+\infty )} [ 0 , + ∞ ) {\displaystyle [0,+\infty )} 反雙曲正切 y = a r t a n h x {\displaystyle y=\mathrm {artanh} x} 1 2 ln ( 1 + x 1 − x ) {\displaystyle {\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)} ( − 1 , 1 ) {\displaystyle (-1,1)} R {\displaystyle \mathbb {R} } 反雙曲餘切 y = a r c o t h x {\displaystyle y=\mathrm {arcoth} x} 1 2 ln ( x + 1 x − 1 ) {\displaystyle {\frac {1}{2}}\ln \left({\frac {x+1}{x-1}}\right)} ( − ∞ , − 1 ) ∪ ( 1 , + ∞ ) {\displaystyle (-\infty ,-1)\cup (1,+\infty )} ( − ∞ , 0 ) ∪ ( 0 , + ∞ ) {\displaystyle (-\infty ,0)\cup (0,+\infty )} 反雙曲正割 y = a r s e c h x {\displaystyle y=\mathrm {arsech} x} ln ( 1 x + 1 − x 2 x ) {\displaystyle \ln \left({\frac {1}{x}}+{\frac {\sqrt {1-x^{2}}}{x}}\right)} ( 0 , 1 ] {\displaystyle (0,1]} [ 0 , + ∞ ) {\displaystyle [0,+\infty )} 反雙曲餘割 y = a r c s c h x {\displaystyle y=\mathrm {arcsch} x} ln ( 1 x + 1 + x 2 | x | ) {\displaystyle \ln \left({\frac {1}{x}}+{\frac {\sqrt {1+x^{2}}}{\left|x\right|}}\right)} ( − ∞ , 0 ) ∪ ( 0 , + ∞ ) {\displaystyle (-\infty ,0)\cup (0,+\infty )} ( − ∞ , 0 ) ∪ ( 0 , + ∞ ) {\displaystyle (-\infty ,0)\cup (0,+\infty )} 關閉 Remove ads d d x arsinh x = 1 1 + x 2 d d x arcosh x = 1 x 2 − 1 , x > 1 d d x artanh x = 1 1 − x 2 , | x | < 1 d d x arcoth x = 1 1 − x 2 , | x | > 1 d d x arsech x = − 1 x 1 − x 2 , x ∈ ( 0 , 1 ) d d x arcsch x = − 1 | x | 1 + x 2 , x ≠ 0 {\displaystyle {\begin{aligned}{\frac {d}{dx}}\operatorname {arsinh} \,x&{}={\frac {1}{\sqrt {1+x^{2}}}}\\{\frac {d}{dx}}\operatorname {arcosh} \,x&{}={\frac {1}{\sqrt {x^{2}-1}}},\qquad x>1\\{\frac {d}{dx}}\operatorname {artanh} \,x&{}={\frac {1}{1-x^{2}}},\qquad |x|<1\\{\frac {d}{dx}}\operatorname {arcoth} \,x&{}={\frac {1}{1-x^{2}}},\qquad |x|>1\\{\frac {d}{dx}}\operatorname {arsech} \,x&{}={\frac {-1}{x{\sqrt {1-x^{2}}}}},\qquad x\in (0,1)\\{\frac {d}{dx}}\operatorname {arcsch} \,x&{}={\frac {-1}{|x|{\sqrt {1+x^{2}}}}},\qquad x{\text{ ≠ }}0\\\end{aligned}}} 求導範例: 設θ = arsinh x,則: d arsinh x d x = d θ d sinh θ = 1 cosh θ = 1 1 + sinh 2 θ = 1 1 + x 2 {\displaystyle {\frac {d\,\operatorname {arsinh} \,x}{dx}}={\frac {d\theta }{d\sinh \theta }}={\frac {1}{\cosh \theta }}={\frac {1}{\sqrt {1+\sinh ^{2}\theta }}}={\frac {1}{\sqrt {1+x^{2}}}}} Remove ads arsinh x {\displaystyle \operatorname {arsinh} \,x} = x − ( 1 2 ) x 3 3 + ( 1 ⋅ 3 2 ⋅ 4 ) x 5 5 − ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) x 7 7 + ⋯ {\displaystyle =x-\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}+\cdots } = ∑ n = 0 ∞ ( ( − 1 ) n ( 2 n ) ! 2 2 n ( n ! ) 2 ) x 2 n + 1 ( 2 n + 1 ) , | x | < 1 {\displaystyle =\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n+1}}{(2n+1)}},\qquad \left|x\right|<1} arcosh x {\displaystyle \operatorname {arcosh} \,x} = ln 2 x − ( ( 1 2 ) x − 2 2 + ( 1 ⋅ 3 2 ⋅ 4 ) x − 4 4 + ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) x − 6 6 + ⋯ ) {\displaystyle =\ln 2x-\left(\left({\frac {1}{2}}\right){\frac {x^{-2}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{-4}}{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{-6}}{6}}+\cdots \right)} = ln 2 x − ∑ n = 1 ∞ ( ( − 1 ) n ( 2 n ) ! 2 2 n ( n ! ) 2 ) x − 2 n ( 2 n ) , x > 1 {\displaystyle =\ln 2x-\sum _{n=1}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{-2n}}{(2n)}},\qquad x>1} artanh x = x + x 3 3 + x 5 5 + x 7 7 + ⋯ = ∑ n = 0 ∞ x 2 n + 1 ( 2 n + 1 ) , | x | < 1 {\displaystyle \operatorname {artanh} \,x=x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+{\frac {x^{7}}{7}}+\cdots =\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{(2n+1)}},\qquad \left|x\right|<1} arcsch x = arsinh x − 1 {\displaystyle \operatorname {arcsch} \,x=\operatorname {arsinh} \,x^{-1}} = x − 1 − ( 1 2 ) x − 3 3 + ( 1 ⋅ 3 2 ⋅ 4 ) x − 5 5 − ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) x − 7 7 + ⋯ {\displaystyle =x^{-1}-\left({\frac {1}{2}}\right){\frac {x^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{-5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{-7}}{7}}+\cdots } = ∑ n = 0 ∞ ( ( − 1 ) n ( 2 n ) ! 2 2 n ( n ! ) 2 ) x − ( 2 n + 1 ) ( 2 n + 1 ) , | x | < 1 {\displaystyle =\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{-(2n+1)}}{(2n+1)}},\qquad \left|x\right|<1} arsech x = arcosh x − 1 {\displaystyle \operatorname {arsech} \,x=\operatorname {arcosh} \,x^{-1}} = ln 2 x − ( ( 1 2 ) x 2 2 + ( 1 ⋅ 3 2 ⋅ 4 ) x 4 4 + ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) x 6 6 + ⋯ ) {\displaystyle =\ln {\frac {2}{x}}-\left(\left({\frac {1}{2}}\right){\frac {x^{2}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{4}}{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{6}}{6}}+\cdots \right)} = ln 2 x − ∑ n = 1 ∞ ( ( − 1 ) n ( 2 n ) ! 2 2 n ( n ! ) 2 ) x 2 n 2 n , 0 < x ≤ 1 {\displaystyle =\ln {\frac {2}{x}}-\sum _{n=1}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n}}{2n}},\qquad 0<x\leq 1} arcoth x = artanh x − 1 {\displaystyle \operatorname {arcoth} \,x=\operatorname {artanh} \,x^{-1}} = x − 1 + x − 3 3 + x − 5 5 + x − 7 7 + ⋯ {\displaystyle =x^{-1}+{\frac {x^{-3}}{3}}+{\frac {x^{-5}}{5}}+{\frac {x^{-7}}{7}}+\cdots } = ∑ n = 0 ∞ x − ( 2 n + 1 ) ( 2 n + 1 ) , | x | > 1 {\displaystyle =\sum _{n=0}^{\infty }{\frac {x^{-(2n+1)}}{(2n+1)}},\qquad \left|x\right|>1} arcosh ( 2 x 2 − 1 ) = 2 arcosh x {\displaystyle \operatorname {arcosh} (2x^{2}-1)=2\operatorname {arcosh} x} arcosh ( 2 x 2 + 1 ) = 2 arsinh x {\displaystyle \operatorname {arcosh} (2x^{2}+1)=2\operatorname {arsinh} x} Remove ads ∫ arsinh x d x = x arsinh x − x 2 + 1 + C ∫ arcosh x d x = x arcosh x − x 2 − 1 + C , x > 1 ∫ artanh x d x = x artanh x + 1 2 ln ( 1 − x 2 ) + C , | x | < 1 ∫ arcoth x d x = x arcoth x + 1 2 ln ( x 2 − 1 ) + C , | x | > 1 ∫ arsech x d x = x arsech x + arcsin x + C , x ∈ ( 0 , 1 ) ∫ arcsch x d x = x arcsch x + | arsinh x | + C , x ≠ 0 {\displaystyle {\begin{aligned}\int \operatorname {arsinh} \,x\,dx&{}=x\,\operatorname {arsinh} \,x-{\sqrt {x^{2}+1}}+C\\\int \operatorname {arcosh} \,x\,dx&{}=x\,\operatorname {arcosh} \,x-{\sqrt {x^{2}-1}}+C,\qquad x>1\\\int \operatorname {artanh} \,x\,dx&{}=x\,\operatorname {artanh} \,x+{\frac {1}{2}}\ln \left(1-x^{2}\right)+C,\qquad |x|<1\\\int \operatorname {arcoth} \,x\,dx&{}=x\,\operatorname {arcoth} \,x+{\frac {1}{2}}\ln \left(x^{2}-1\right)+C,\qquad |x|>1\\\int \operatorname {arsech} \,x\,dx&{}=x\,\operatorname {arsech} \,x+\arcsin \,x+C,\qquad x\in (0,1)\\\int \operatorname {arcsch} \,x\,dx&{}=x\,\operatorname {arcsch} \,x+\left|\operatorname {arsinh} \,x\right|+C,\qquad x\neq 0\end{aligned}}} 使用分部積分法和上面的簡單導數很容易得出它們。 Remove ads [註 1]雙曲餘弦函數是偶函數,所以對於一個y值(y>1),都有兩個x值與之對應,取反的時候只取一個(通常是正的)即可。 Inverse trigonometric functions(頁面存檔備份,存於互聯網檔案館) at MathWorld 雙曲函數Wikiwand in your browser!Seamless Wikipedia browsing. 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