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Math formulas: Hyperbolic functions
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Definitions of hyperbolic functions

$$ \sinh x=\frac{e^x - e^{-x}}{2} $$
$$ \cosh x=\frac{e^x + e^{-x}}{2} $$
$$ \tanh x=\frac{e^x - e^{-x}}{e^x + e^{-x}} =\frac{\sinh x}{\cosh x} $$
$$ \mathrm{csch}\,x=\frac{2}{e^x - e^{-x}} = \frac{1}{\sinh x} $$
$$ \mathrm{sech}\,x=\frac{2}{e^x + e^{-x}} = \frac{1}{\cosh x} $$
$$ \coth\,x=\frac{e^x + e^{-x}}{e^x - e^{-x}} = \frac{\cosh x}{\sinh x} $$

Derivatives

$$ \frac{d}{dx}\, \sinh x = \cosh x $$
$$ \frac{d}{dx}\, \cosh x = \sinh x $$
$$ \frac{d}{dx}\, \tanh x = \mathrm{sech}^2x $$
$$ \frac{d}{dx}\, \mathrm{csch}\,x = -\mathrm{csch}\,x\cdot \coth x $$
$$ \frac{d}{dx}\, \mathrm{sech}\,x = -\mathrm{sech}\,x\cdot \tanh x $$
$$ \frac{d}{dx}\,\coth x = -\mathrm{csch}^2x $$

Hyperbolic identities

$$ \cosh^2x - \sinh^2x = 1 $$
$$ \tanh^2x + \mathrm{sech}^2x = 1 $$
$$ \coth^2x - \mathrm{csch}^2x = 1 $$
$$ \sinh(x \pm y) = \sinh x \cdot \cosh y \pm \cosh x\cdot \sinh y $$
$$ \cosh(x \pm y) = \cosh x \cdot \cosh y \pm \sinh x \cdot \sinh y $$
$$ \sinh(2\cdot x) = 2 \cdot \sinh x \cdot \cosh x $$
$$ \cosh(2\cdot x) = \cosh^2x + \sinh^2x $$
$$ \sinh^2x = \frac{-1 + \cosh 2x}{2} $$
$$ \cosh^2x = \frac{1 + \cosh 2x}{2} $$

Inverse Hyperbolic functions

$$ \sinh^{-1}x=\ln \left(x+\sqrt{x^2 + 1}\right), ~~ x \in (-\infty, \infty) $$
$$ \cosh^{-1}x=\ln\left(x+\sqrt{x^2 - 1}\right), ~~ x \in [1, \infty) $$
$$ \tanh^{-1}x=\frac{1}{2} \ln\left(\frac{1 + x}{1 -x}\right), ~~ x \in (-1, 1) $$
$$ \coth^{-1}x=\frac{1}{2}\,\ln\left(\frac{x + 1}{x-1}\right), ~~ x \in (-\infty, -1) \cup (1, \infty) $$
$$ \mathrm{sech}^{-1}x=\ln\left(\frac{1 + \sqrt{1-x^2}}{x}\right), ~~ x \in (0, 1]$$
$$ \mathrm{csch}^{-1}x = \ln\left(\frac{1}{x} + \frac{\sqrt{1-x^2}}{|x|}\right), ~~ x \in (-\infty, 0) \cup (0,\infty)$$

Derivatives of Inverse Hyperbolic functions

$$ \frac{d}{dx}\,\sinh^{-1}x= \frac{1}{\sqrt{x^2+1}} $$
$$ \frac{d}{dx}\, \cosh^{-1}x=\frac{1}{\sqrt{x^2-1}} $$
$$ \frac{d}{dx}\,tanh^{-1}x=\frac{1}{1-x^2} $$
$$ \frac{d}{dx}\, \mathrm{csch}^{-1}x=-\frac{1}{|x|\sqrt{1 + x^2}} $$
$$ \frac{d}{dx}\,\mathrm{sech}^{-1}x=-\frac{1}{x\sqrt{1 - x^2}} $$
$$ \frac{d}{dx}\,\coth^{-1}x=\frac{1}{1-x^2} $$
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