# Theory

## Hyperbolic Functions

### Definitions

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

$\sinh(x)={\frac {e^{x}-e^{-x}}{2}}$ $\cosh(x)={\frac {e^{x}+e^{-x}}{2}}$ $\tanh(x)={\frac {\sinh(x)}{\cosh(x)}}={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}$ The reciprocal functions csch, sech, coth are defined from these functions:

${\rm {csch}}(x)={\frac {1}{\sinh(x)}}$ ${\rm {sech}}(x)={\frac {1}{\cosh(x)}}$ $\coth(x)={\frac {1}{\tanh(x)}}$ ### Some simple identities

$\cosh ^{2}(x)-\sinh ^{2}(x)=1$ $1-\tanh ^{2}(x)={\rm {sech}}^{2}(x)$ $\sinh(2x)=2\sinh(x)\cosh(x)$ $\cosh(2x)=\cosh ^{2}(x)+\sinh ^{2}(x)$ ### Derivatives of hyperbolic functions

${\frac {d}{dx}}\sinh(x)=\cosh(x)$ ${\frac {d}{dx}}\cosh(x)=\sinh(x)$ ${\frac {d}{dx}}\tanh(x)={\rm {sech}}^{2}(x)$ ${\frac {d}{dx}}{\rm {csch}}(x)=-{\rm {csch}}(x){\rm {coth}}(x)$ ${\frac {d}{dx}}{\rm {sech}}(x)=-{\rm {sech}}(x)\tanh(x)$ ${\frac {d}{dx}}{\rm {coth}}(x)=-{\rm {csch}}^{2}(x)$ ### 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.

$\sinh :\mathbb {R} \to \mathbb {R}$ $\cosh :[0,\infty ]\to [1,\infty ]$ $\tanh :\mathbb {R} \to (-1,1)$ ### Inverse hyperbolic functions

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

${\rm {arsinh:\mathbb {R} \to \mathbb {R} }}$ ${\rm {arcosh:[1,\infty ]\to [0,\infty ]}}$ ${\rm {artanh:(-1,1)\to \mathbb {R} }}$ We can define ${\rm {arcsch}}$ , ${\rm {arsech}}$ and ${\rm {arcoth}}$ similarly.

We can also write these inverses using the logarithm function,

${\rm {arsinh}}(x)=\ln {\big (}x+{\sqrt {x^{2}+1}}{\big )}$ ${\rm {arcosh}}(x)=\ln {\big (}x+{\sqrt {x^{2}-1}}{\big )}$ ${\rm {artanh}}(x)=\ln \left({\sqrt {\frac {1+x}{1-x}}}\right)$ These identities can simplify some integrals.

### Derivatives of inverse hyperbolic functions

${\frac {d}{dx}}{\rm {arsinh}}(x)={\frac {1}{\sqrt {1+x^{2}}}}$ ${\frac {d}{dx}}{\rm {arcosh}}(x)={\frac {1}{\sqrt {x^{2}-1}}}\ ,\ x>1$ ${\frac {d}{dx}}{\rm {artanh}}(x)={\frac {1}{1-x^{2}}}\ ,\ |x|<1$ ${\frac {d}{dx}}{\rm {arcsch}}(x)=-{\frac {1}{|x|{\sqrt {1+x^{2}}}}}\ ,\ x\neq 0$ ${\frac {d}{dx}}{\rm {arsech}}(x)=-{\frac {1}{x{\sqrt {1-x^{2}}}}}\ ,\ 0 ${\frac {d}{dx}}{\rm {arcoth}}(x)={\frac {1}{1-x^{2}}}\ ,\ |x|>1$ ## Transcendental Functions

Transcendental functions are not algebraic. These include trigonometric, inverse trigonometric, logarithmic and exponential functions and many others.