Fungsi hiperbolik adalah salah satu hasil kombinasi dari fungsi-fungsi eksponen. Fungsi hiperbolik memiliki rumus. Selain itu memiliki invers serta turunan dan anti turunan fungsi hiperbolik dan inversnya.^[1]

Dalam istilah dari fungsi eksponensial:

• Hiperbolik sinus:

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

• Hiperbolik kosinus:

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

• Hiperbolik tangen:

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

• Hiperbolik kotangen: untuk x ≠ 0,

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

• Hiperbolik sekan:

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

• Hiperbolik kosekan: untuk x ≠ 0,

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

- Dalam pengembangan -

-Dalam pengembangan -

- Dalam pengembangan -

${\displaystyle {\begin{aligned}\sinh(x+y)&=\sinh x\cosh y+\cosh x\sinh y\\\cosh(x+y)&=\cosh x\cosh y+\sinh x\sinh y\\[6px]\tanh(x+y)&={\frac {\tanh x+\tanh y}{1+\tanh x\tanh y}}\\\end{aligned}}}$

terutama

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

Lihat:

${\displaystyle {\begin{aligned}\sinh x+\sinh y&=2\sinh \left({\frac {x+y}{2}}\right)\cosh \left({\frac {x-y}{2}}\right)\\\cosh x+\cosh y&=2\cosh \left({\frac {x+y}{2}}\right)\cosh \left({\frac {x-y}{2}}\right)\\\end{aligned}}}$

${\displaystyle {\begin{aligned}\sinh(x-y)&=\sinh x\cosh y-\cosh x\sinh y\\\cosh(x-y)&=\cosh x\cosh y-\sinh x\sinh y\\\tanh(x-y)&={\frac {\tanh x-\tanh y}{1-\tanh x\tanh y}}\\\end{aligned}}}$

Dan juga:^[2]

${\displaystyle {\begin{aligned}\sinh x-\sinh y&=2\cosh \left({\frac {x+y}{2}}\right)\sinh \left({\frac {x-y}{2}}\right)\\\cosh x-\cosh y&=2\sinh \left({\frac {x+y}{2}}\right)\sinh \left({\frac {x-y}{2}}\right)\\\end{aligned}}}$

${\displaystyle {\begin{aligned}\sinh \left({\frac {x}{2}}\right)&={\frac {\sinh x}{\sqrt {2(\cosh x+1)}}}&&=\operatorname {sgn} x\,{\sqrt {\frac {\cosh x-1}{2}}}\\[6px]\cosh \left({\frac {x}{2}}\right)&={\sqrt {\frac {\cosh x+1}{2}}}\\[6px]\tanh \left({\frac {x}{2}}\right)&={\frac {\sinh x}{\cosh x+1}}&&=\operatorname {sgn} x\,{\sqrt {\frac {\cosh x-1}{\cosh x+1}}}={\frac {e^{x}-1}{e^{x}+1}}\end{aligned}}}$

di mana sgn adalah fungsi tanda.

Jika ${\displaystyle x\neq 0}$, maka^[3]

${\displaystyle \tanh \left({\frac {x}{2}}\right)={\frac {\cosh x-1}{\sinh x}}=\coth x-\operatorname {csch} x}$

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

Pertidaksamaan berikut sangat berguna dalam statistik, yaitu ${\displaystyle \operatorname {cosh} (t)\leq e^{t^{2}/2}}$^[4]

${\displaystyle {\begin{aligned}\operatorname {arsinh} (x)&=\ln \left(x+{\sqrt {x^{2}+1}}\right)\\\operatorname {arcosh} (x)&=\ln \left(x+{\sqrt {x^{2}-1}}\right)&&x\geqslant 1\\\operatorname {artanh} (x)&={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)&&|x|<1\\\operatorname {arcoth} (x)&={\frac {1}{2}}\ln \left({\frac {x+1}{x-1}}\right)&&|x|>1\\\operatorname {arsech} (x)&=\ln \left({\frac {1}{x}}+{\sqrt {{\frac {1}{x^{2}}}-1}}\right)=\ln \left({\frac {1+{\sqrt {1-x^{2}}}}{x}}\right)&&0<x\leqslant 1\\\operatorname {arcsch} (x)&=\ln \left({\frac {1}{x}}+{\sqrt {{\frac {1}{x^{2}}}+1}}\right)=\ln \left({\frac {1+{\sqrt {1+x^{2}}}}{x}}\right)&&x\neq 0\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {d}{dx}}\sinh x&=\cosh x\\{\frac {d}{dx}}\cosh x&=\sinh x\\{\frac {d}{dx}}\tanh x&=1-\tanh ^{2}x=\operatorname {sech} ^{2}x={\frac {1}{\cosh ^{2}x}}\\{\frac {d}{dx}}\coth x&=1-\coth ^{2}x=-\operatorname {csch} ^{2}x=-{\frac {1}{\sinh ^{2}x}}&&x\neq 0\\{\frac {d}{dx}}\operatorname {sech} x&=-\tanh x\operatorname {sech} x\\{\frac {d}{dx}}\operatorname {csch} x&=-\coth x\operatorname {csch} x&&x\neq 0\\{\frac {d}{dx}}\operatorname {arsinh} x&={\frac {1}{\sqrt {x^{2}+1}}}\\{\frac {d}{dx}}\operatorname {arcosh} x&={\frac {1}{\sqrt {x^{2}-1}}}&&1<x\\{\frac {d}{dx}}\operatorname {artanh} x&={\frac {1}{1-x^{2}}}&&|x|<1\\{\frac {d}{dx}}\operatorname {arcoth} x&={\frac {1}{1-x^{2}}}&&1<|x|\\{\frac {d}{dx}}\operatorname {arsech} x&=-{\frac {1}{x{\sqrt {1-x^{2}}}}}&&0<x<1\\{\frac {d}{dx}}\operatorname {arcsch} x&=-{\frac {1}{|x|{\sqrt {1+x^{2}}}}}&&x\neq 0\end{aligned}}}$

- Dalam pengembangan -

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

${\displaystyle {\begin{aligned}\int {{\frac {1}{\sqrt {a^{2}+u^{2}}}}\,du}&=\operatorname {arsinh} \left({\frac {u}{a}}\right)+C\\\int {{\frac {1}{\sqrt {u^{2}-a^{2}}}}\,du}&=\operatorname {arcosh} \left({\frac {u}{a}}\right)+C\\\int {\frac {1}{a^{2}-u^{2}}}\,du&=a^{-1}\operatorname {artanh} \left({\frac {u}{a}}\right)+C&&u^{2}<a^{2}\\\int {\frac {1}{a^{2}-u^{2}}}\,du&=a^{-1}\operatorname {arcoth} \left({\frac {u}{a}}\right)+C&&u^{2}>a^{2}\\\int {{\frac {1}{u{\sqrt {a^{2}-u^{2}}}}}\,du}&=-a^{-1}\operatorname {arsech} \left({\frac {u}{a}}\right)+C\\\int {{\frac {1}{u{\sqrt {a^{2}+u^{2}}}}}\,du}&=-a^{-1}\operatorname {arcsch} \left|{\frac {u}{a}}\right|+C\end{aligned}}}$

1. ^ "FUNGSI HIPERBOLIK DAN INVERSNYA". DIGILIB UNNES. Diarsipkan dari versi asli tanggal 2019-08-15. Diakses tanggal 2014-05-28. 
2. ^ Martin, George E. (1986). The foundations of geometry and the non-euclidean plane (edisi ke-1st corr.). New York: Springer-Verlag. hlm. 416. ISBN 3-540-90694-0. 
3. ^ "Prove the identity". StackExchange (mathematics). Diarsipkan dari versi asli tanggal 2023-07-26. Diakses tanggal 24 January 2016. 
4. ^ Audibert, Jean-Yves (2009). "Fast learning rates in statistical inference through aggregation". The Annals of Statistics. hlm. 1627.  [1] Diarsipkan 2023-07-26 di Wayback Machine.