Theory

The unit hyperbola has a sector with an area half of the hyperbolic angle

The independent variable of a hyperbolic function is called a hyperbolic angle. Just as the circular functions sine and cosine can be seen as projections from the unit circle to the axes, so the hyperbolic functions sinh and cosh are projections from a unit hyperbola to the axes.

Definitions

The hyperbolic functions are defined in analogy with the trigonometric functions:

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

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

${\displaystyle \tanh(x)={\frac {\sinh(x)}{\cosh(x)}}={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}}$

The reciprocal functions csch, sech, coth are defined from these functions:

${\displaystyle {\rm {csch}}(x)={\frac {1}{\sinh(x)}}}$

${\displaystyle {\rm {sech}}(x)={\frac {1}{\cosh(x)}}}$

${\displaystyle \coth(x)={\frac {1}{\tanh(x)}}}$

Some simple identities

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

${\displaystyle 1-\tanh ^{2}(x)={\rm {sech}}^{2}(x)}$

${\displaystyle \sinh(2x)=2\sinh(x)\cosh(x)}$

${\displaystyle \cosh(2x)=\cosh ^{2}(x)+\sinh ^{2}(x)}$

Derivatives of hyperbolic functions

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

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

${\displaystyle {\frac {d}{dx}}\tanh(x)={\rm {sech}}^{2}(x)}$

${\displaystyle {\frac {d}{dx}}{\rm {csch}}(x)=-{\rm {csch}}(x){\rm {coth}}(x)}$

${\displaystyle {\frac {d}{dx}}{\rm {sech}}(x)=-{\rm {sech}}(x)\tanh(x)}$

${\displaystyle {\frac {d}{dx}}{\rm {coth}}(x)=-{\rm {csch}}^{2}(x)}$

Principal values of the main hyperbolic functions

There is no problem in defining principal braches for sinh and tanh because they are injective. We choose one of the principal branches for cosh.

${\displaystyle \sinh :\mathbb {R} \to \mathbb {R} }$

${\displaystyle \cosh :[0,\infty ]\to [1,\infty ]}$

${\displaystyle \tanh :\mathbb {R} \to (-1,1)}$

Inverse hyperbolic functions

With the principal values defined above, the definition of the inverse functions is immediate:

${\displaystyle {\rm {arsinh:\mathbb {R} \to \mathbb {R} }}}$

${\displaystyle {\rm {arcosh:[1,\infty ]\to [0,\infty ]}}}$

${\displaystyle {\rm {artanh:(-1,1)\to \mathbb {R} }}}$

We can define ${\displaystyle {\rm {arcsch}}}$ , ${\displaystyle {\rm {arsech}}}$ and ${\displaystyle {\rm {arcoth}}}$ similarly.

We can also write these inverses using the logarithm function,

${\displaystyle {\rm {arsinh}}(x)=\ln {\big (}x+{\sqrt {x^{2}+1}}{\big )}}$

${\displaystyle {\rm {arcosh}}(x)=\ln {\big (}x+{\sqrt {x^{2}-1}}{\big )}}$

${\displaystyle {\rm {artanh}}(x)=\ln \left({\sqrt {\frac {1+x}{1-x}}}\right)}$

These identities can simplify some integrals.

Derivatives of inverse hyperbolic functions

${\displaystyle {\frac {d}{dx}}{\rm {arsinh}}(x)={\frac {1}{\sqrt {1+x^{2}}}}}$

${\displaystyle {\frac {d}{dx}}{\rm {arcosh}}(x)={\frac {1}{\sqrt {x^{2}-1}}}\ ,\ x>1}$

${\displaystyle {\frac {d}{dx}}{\rm {artanh}}(x)={\frac {1}{1-x^{2}}}\ ,\ |x|<1}$

${\displaystyle {\frac {d}{dx}}{\rm {arcsch}}(x)=-{\frac {1}{|x|{\sqrt {1+x^{2}}}}}\ ,\ x\neq 0}$

${\displaystyle {\frac {d}{dx}}{\rm {arsech}}(x)=-{\frac {1}{x{\sqrt {1-x^{2}}}}}\ ,\ 0<x<1}$

${\displaystyle {\frac {d}{dx}}{\rm {arcoth}}(x)={\frac {1}{1-x^{2}}}\ ,\ |x|>1}$

Transcendental Functions

Hyperbolic functions are examples of transcendental functions -- they are not algebraic functions. They include trigonometric, inverse trigonometric, logarithmic and exponential functions.