تابع هذلولوی

توابع هایپربولیک از این قراراند:

${\displaystyle \sinh x={\frac {e^{x}-e^{-x}}{2}}}$

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

${\displaystyle \tanh x={\frac {\sinh x}{\cosh x}}={\frac {\frac {e^{x}-e^{-x}}{2}}{\frac {e^{x}+e^{-x}}{2}}}={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}={\frac {e^{2x}-1}{e^{2x}+1}}}$

${\displaystyle \coth x={\frac {\cosh x}{\sinh x}}={\frac {\frac {e^{x}+e^{-x}}{2}}{\frac {e^{x}-e^{-x}}{2}}}={\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}={\frac {e^{2x}+1}{e^{2x}-1}}}$

${\displaystyle \operatorname {sech} x={\frac {1}{\cosh x}}={\frac {2}{e^{x}+e^{-x}}}}$

${\displaystyle \operatorname {csch} x={\frac {1}{\sinh x}}={\frac {2}{e^{x}-e^{-x}}}}$

رابطهٔ توابع هایپربولیک با توابع مثلثلتی چنین است:

${\displaystyle \sinh x=-{\rm {i}}\sin {\rm {i}}x\!}$

${\displaystyle \cosh x=\cos {\rm {i}}x\!}$

${\displaystyle \tanh x=-{\rm {i}}\tan {\rm {i}}x\!}$

${\displaystyle \coth x={\rm {i}}\cot {\rm {i}}x\!}$

${\displaystyle \operatorname {sech} \,x=\sec {{\rm {i}}x}\!}$

${\displaystyle \operatorname {csch} \,x={\rm {i}}\,\csc \,{\rm {i}}x\!}$

که در آن i یکهٔ موهومی با تعریف i = −۱ است.

${\displaystyle coshx}$

${\displaystyle {\begin{aligned}\sinh(-x)&=-\sinh x\\\cosh(-x)&=\cosh x\\\tanh(-x)&=-\tanh x\\\coth(-x)&=-\coth x\\\operatorname {sech} (-x)&=\operatorname {sech} x\\\operatorname {csch} (-x)&=-\operatorname {csch} x\end{aligned}}}$

همچنین داریم:

${\displaystyle {\begin{aligned}\operatorname {arsech} x&=\operatorname {arcosh} {\frac {1}{x}}\\\operatorname {arcsch} x&=\operatorname {arsinh} {\frac {1}{x}}\\\operatorname {arcoth} x&=\operatorname {artanh} {\frac {1}{x}}\end{aligned}}}$

متناظر با روابط مثلثاتی داریم:

${\textstyle {\begin{aligned}\cosh ^{2}x-\sinh ^{2}x=1\\\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {sech} ^{2}x&=1-\tanh ^{2}x\\\operatorname {csch} ^{2}x&=\coth ^{2}x-1\\\end{aligned}}}$

مجموع دو عبارت:

${\displaystyle {\begin{aligned}\cosh(x+y)&=\sinh x\sinh y+\cosh x\cosh y\\\sinh(x+y)&=\cosh x\sinh y+\sinh x\cosh y\end{aligned}}}$

مشخصاً

${\displaystyle {\begin{aligned}\cosh(2x)&=\sinh ^{2}{x}+\cosh ^{2}{x}=2\sinh ^{2}x+1=2\cosh ^{2}x-1\\\sinh(2x)&=2\sinh x\cosh x\end{aligned}}}$

مجموع و تفاضل ${\displaystyle coshx}$

${\displaystyle {\begin{aligned}\cosh x+\sinh x&=e^{x}\\\cosh x-\sinh x&=e^{-x}\end{aligned}}}$

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

${\displaystyle arccoshx=\cosh ^{-1}x=\ln \left(x+{\sqrt {x^{2}-1}}\right);x\geq 1}$

${\displaystyle arctanhx=\tanh ^{-1}x={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right);\left|x\right|<1}$

${\displaystyle {\frac {d}{dx}}\sinh x=\cosh x\,}$

${\displaystyle {\frac {d}{dx}}\cosh x=\sinh x\,}$

${\displaystyle {\frac {d}{dx}}\tanh x=1-\tanh ^{2}x=\operatorname {sech} ^{2}x=1/\cosh ^{2}x\,}$

${\displaystyle {\frac {d}{dx}}\coth x=1-\coth ^{2}x=-\operatorname {csch} ^{2}x=-1/\sinh ^{2}x\,}$

${\displaystyle {\frac {d}{dx}}\ \operatorname {csch} \,x=-\coth x\ \operatorname {csch} \,x\,}$

${\displaystyle {\frac {d}{dx}}\ \operatorname {sech} \,x=-\tanh x\ \operatorname {sech} \,x\,}$

${\displaystyle {\frac {d}{dx}}\,\operatorname {arsinh} \,x={\frac {1}{\sqrt {x^{2}+1}}}}$

${\displaystyle {\frac {d}{dx}}\,\operatorname {arcosh} \,x={\frac {1}{\sqrt {x^{2}-1}}}}$

${\displaystyle {\frac {d}{dx}}\,\operatorname {artanh} \,x={\frac {1}{1-x^{2}}}}$

${\displaystyle {\frac {d}{dx}}\,\operatorname {arcsch} \,x=-{\frac {1}{\left|x\right|{\sqrt {1+x^{2}}}}}}$

${\displaystyle {\frac {d}{dx}}\,\operatorname {arsech} \,x=-{\frac {1}{x{\sqrt {1-x^{2}}}}}}$

${\displaystyle {\frac {d}{dx}}\,\operatorname {arcoth} \,x={\frac {1}{1-x^{2}}}}$

برای فهرست کاملی از این انتگرالها، فهرست انتگرال‌های تابع‌های هیپربولیک را ببینید.

${\displaystyle {\begin{aligned}\int \sinh(ax)\,dx&=a^{-1}\cosh(ax)+C\\\int \cosh(ax)\,dx&=a^{-1}\sinh(ax)+C\\\int \tanh(ax)\,dx&=a^{-1}\ln(\cosh(ax))+C\\\int \coth(ax)\,dx&=a^{-1}\ln(\sinh(ax))+C\\\int \operatorname {sech} (ax)\,dx&=a^{-1}\arctan(\sinh(ax))+C\\\int \operatorname {csch} (ax)\,dx&=a^{-1}\ln \left(\tanh \left({\frac {ax}{2}}\right)\right)+C\end{aligned}}}$

${\displaystyle {\begin{aligned}\int {\frac {du}{\sqrt {a^{2}+u^{2}}}}&=\sinh ^{-1}\left({\frac {u}{a}}\right)+C\\\int {\frac {du}{\sqrt {u^{2}-a^{2}}}}&=\cosh ^{-1}\left({\frac {u}{a}}\right)+C\\\int {\frac {du}{a^{2}-u^{2}}}&=a^{-1}\tanh ^{-1}\left({\frac {u}{a}}\right)+C;u^{2}<a^{2}\\\int {\frac {du}{a^{2}-u^{2}}}&=a^{-1}\coth ^{-1}\left({\frac {u}{a}}\right)+C;u^{2}>a^{2}\\\int {\frac {du}{u{\sqrt {a^{2}-u^{2}}}}}&=-a^{-1}\operatorname {sech} ^{-1}\left({\frac {u}{a}}\right)+C\\\int {\frac {du}{u{\sqrt {a^{2}+u^{2}}}}}&=-a^{-1}\operatorname {csch} ^{-1}\left|{\frac {u}{a}}\right|+C\end{aligned}}}$

• معکوس توابع هیپربولیک