## 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:

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

## Derivatives of Inverse Hyperbolic functions:

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