دوال زائدية عكسية

${\displaystyle \forall x\in [1,+\infty [\;:\;\operatorname {arcosh} (x)=\ln(x+{\sqrt {x^{2}-1}})}$

${\displaystyle {\begin{aligned}y&=\operatorname {arcosh} (x);\quad x\geq 1\\x&=\operatorname {cosh} (y);\quad \ \ \ \ y\geq 0\end{aligned}}}$

${\displaystyle \cosh(y)+\sinh(y)=e^{y}....(*)}$

${\displaystyle \cosh ^{2}(y)-\sinh ^{2}(y)=1}$

${\displaystyle \sinh ^{2}(y)=\cosh ^{2}(y)-1}$

${\displaystyle {\begin{aligned}|\sinh(y)|&={\sqrt {\cosh ^{2}(y)-1}}\;;\;y\geq 0\Rightarrow \sinh(y)\geq 0\\\sinh(y)&={\sqrt {\cosh ^{2}(y)-1}}\\(*)&:\cosh(y)+{\sqrt {\cosh ^{2}(y)-1}}=e^{y}\\\Rightarrow y&=\ln \left(\cosh(y)+{\sqrt {\cosh ^{2}(y)-1}}\right)\\\operatorname {arcosh} (x)&=\ln \left(x+{\sqrt {x^{2}-1}}\right)\end{aligned}}}$

${\displaystyle x=\cosh y={\frac {e^{y}+e^{-y}}{2}}}$

${\displaystyle 2x={e^{y}+e^{-y}}}$

${\displaystyle e^{y}-2x+e^{-y}=0}$

${\displaystyle e^{y}-2x+{\frac {1}{e^{y}}}=0}$

${\displaystyle e^{2y}-2xe^{y}+1=0}$

${\displaystyle u^{2}-2xu+1=0}$

${\displaystyle u={\frac {2x\pm {\sqrt {4x^{2}-4}}}{2}}}$

${\displaystyle e^{y}={\frac {2x\pm {\sqrt {4x^{2}-4}}}{2}}}$

${\displaystyle e^{y}={\frac {2x\pm {\sqrt {4(x^{2}-1)}}}{2}}}$

${\displaystyle 2e^{y}={2x\pm 2{\sqrt {x^{2}-1}}}}$

${\displaystyle e^{y}={x\pm {\sqrt {x^{2}-1}}}}$

${\displaystyle \ln e^{y}=\ln \left(x\pm {\sqrt {x^{2}-1}}\right)}$

${\displaystyle y=\ln \left(x+{\sqrt {x^{2}-1}}\right)}$

${\displaystyle \operatorname {arcosh} x=\ln \left(x+{\sqrt {x^{2}-1}}\right)}$

${\displaystyle \operatorname {arsinh} u\pm \operatorname {arsinh} v=\operatorname {arsinh} \left(u{\sqrt {1+v^{2}}}\pm v{\sqrt {1+u^{2}}}\right)}$

${\displaystyle \operatorname {arcosh} u\pm \operatorname {arcosh} v=\operatorname {arcosh} \left(uv\pm {\sqrt {(u^{2}-1)(v^{2}-1)}}\right)}$

${\displaystyle \operatorname {artanh} u\pm \operatorname {artanh} v=\operatorname {artanh} \left({\frac {u\pm v}{1\pm uv}}\right)}$

${\displaystyle {\begin{aligned}\operatorname {arsinh} u+\operatorname {arcosh} v&=\operatorname {arsinh} \left(uv+{\sqrt {(1+u^{2})(v^{2}-1)}}\right)\\&=\operatorname {arcosh} \left(v{\sqrt {1+u^{2}}}+u{\sqrt {v^{2}-1}}\right)\end{aligned}}}$

${\displaystyle {\begin{aligned}&\sinh(\operatorname {arcosh} x)={\sqrt {x^{2}-1}}\quad {\text{for}}\quad |x|>1\\&\sinh(\operatorname {artanh} x)={\frac {x}{\sqrt {1-x^{2}}}}\quad {\text{for}}\quad -1<x<1\\&\cosh(\operatorname {arsinh} x)={\sqrt {1+x^{2}}}\\&\cosh(\operatorname {artanh} x)={\frac {1}{\sqrt {1-x^{2}}}}\quad {\text{for}}\quad -1<x<1\\&\tanh(\operatorname {arsinh} x)={\frac {x}{\sqrt {1+x^{2}}}}\\&\tanh(\operatorname {arcosh} x)={\frac {\sqrt {x^{2}-1}}{x}}\quad {\text{for}}\quad |x|>1\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {d}{dx}}\operatorname {arsinh} x&{}={\frac {1}{\sqrt {x^{2}+1}}},{\text{ for all real }}x\\{\frac {d}{dx}}\operatorname {arcosh} x&{}={\frac {1}{\sqrt {x^{2}-1}}},{\text{ for all real }}x>1\\{\frac {d}{dx}}\operatorname {artanh} x&{}={\frac {1}{1-x^{2}}},{\text{ for all real }}|x|<1\\{\frac {d}{dx}}\operatorname {arcoth} x&{}={\frac {1}{1-x^{2}}},{\text{ for all real }}|x|>1\\{\frac {d}{dx}}\operatorname {arsech} x&{}={\frac {-1}{x{\sqrt {1-x^{2}}}}},{\text{ for all real }}x\in (0,1)\\{\frac {d}{dx}}\operatorname {arcsch} x&{}={\frac {-1}{|x|{\sqrt {1+x^{2}}}}},{\text{ for all real }}x{\text{, except }}0\\\end{aligned}}}$

${\displaystyle \int \operatorname {arsinh} (x)\,dx=x\operatorname {arsinh} (x)-{\sqrt {x^{2}+1}}+C}$

${\displaystyle \int \operatorname {arcosh} (x)\,dx=x\operatorname {arcosh} (x)-{\sqrt {x^{2}-1}}+C}$

${\displaystyle \int \operatorname {artanh} (x)\,dx=x\operatorname {artanh} (x)+{\frac {1}{2}}\ln \left(1-x^{2}\right)+C}$

${\displaystyle \int \operatorname {arcoth} (x)\,dx=x\operatorname {arcoth} (x)+{\frac {1}{2}}\ln \left(x^{2}-1\right)+C}$

${\displaystyle \int \operatorname {arsech} (x)\,dx=x\operatorname {arsech} (x)-{2}\operatorname {arctan} {\sqrt {\frac {1-x}{1+x}}}+C}$

${\displaystyle \int \operatorname {arcsch} (x)\,dx=x\operatorname {arcsch} (x)+\operatorname {arcoth} {\sqrt {{\frac {1}{x^{2}}}+1}}+C}$

${\displaystyle {\begin{aligned}\operatorname {arsinh} x&=x-\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}\pm \cdots \\&=\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n+1}}{2n+1}},\qquad \left|x\right|<1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arcosh} x&=\ln(2x)-\left(\left({\frac {1}{2}}\right){\frac {x^{-2}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{-4}}{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{-6}}{6}}+\cdots \right)\\&=\ln(2x)-\sum _{n=1}^{\infty }\left({\frac {(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{-2n}}{2n}},\qquad \left|x\right|>1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {artanh} x&=x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+{\frac {x^{7}}{7}}+\cdots \\&=\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{2n+1}},\qquad \left|x\right|<1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arcsch} x=\operatorname {arsinh} {\frac {1}{x}}&=x^{-1}-\left({\frac {1}{2}}\right){\frac {x^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{-5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{-7}}{7}}\pm \cdots \\&=\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{-(2n+1)}}{2n+1}},\qquad \left|x\right|>1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arsech} x=\operatorname {arcosh} {\frac {1}{x}}&=\ln {\frac {2}{x}}-\left(\left({\frac {1}{2}}\right){\frac {x^{2}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{4}}{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{6}}{6}}+\cdots \right)\\&=\ln {\frac {2}{x}}-\sum _{n=1}^{\infty }\left({\frac {(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n}}{2n}},\qquad 0<x\leq 1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arcoth} x=\operatorname {artanh} {\frac {1}{x}}&=x^{-1}+{\frac {x^{-3}}{3}}+{\frac {x^{-5}}{5}}+{\frac {x^{-7}}{7}}+\cdots \\&=\sum _{n=0}^{\infty }{\frac {x^{-(2n+1)}}{2n+1}},\qquad \left|x\right|>1\end{aligned}}}$