The hyperbolic sine function is a one-to-one function and thus has an inverse. As usual, the graph of the inverse hyperbolic sine function $sinh^{-1}(x)$ also denoted by $arcsinh(x)$ by reflecting the graph of $sinh(x)$ about the line $y=x$

For all inverse hyperbolic functions but the inverse hyperbolic cotangent and the inverse hyperbolic cosecant, the domain of the real function is connected.

\[\large arcsinh\;x=ln(x+\sqrt {x^{2}+1}\]

\[\large arccosh\;x=ln(x+\sqrt{x^{2}-1})\]

\[\large arctanh\;x=\frac{1}{2}\;ln\left(\frac{1+x}{1-x} \right )\]

\[\large arccoth\;x=\frac{1}{2}\;ln\left(\frac{x+1}{x-1} \right )\]

For all inverse hyperbolic functions but the inverse hyperbolic cotangent and the inverse hyperbolic cosecant, the domain of the real function is connected.

**Inverse hyperbolic sine (if the domain is the whole real line)**\[\large arcsinh\;x=ln(x+\sqrt {x^{2}+1}\]

**Inverse hyperbolic cosine (if the domain is the closed interval $(1, +\infty )$.**\[\large arccosh\;x=ln(x+\sqrt{x^{2}-1})\]

**Inverse hyperbolic tangent [if the domain is the open interval (âˆ’1, 1)]**\[\large arctanh\;x=\frac{1}{2}\;ln\left(\frac{1+x}{1-x} \right )\]

**Inverse hyperbolic cotangent [if the domain is the union of the open intervals (âˆ’âˆž, âˆ’1) and (1, +âˆž)]**\[\large arccoth\;x=\frac{1}{2}\;ln\left(\frac{x+1}{x-1} \right )\]

**Inverse hyperbolic cosecant (if the domain is the real line with 0 removed)**$\large arccsch\;x=ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^{2}}+1}\right)$

**Inverse hyperbolic secant (if the domain is the semi-open intervalÂ 0, 1)**

$\large arcsech\;x=ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^{2}}-1}\right)=ln\left(\frac{1+\sqrt{1-x^{2}}}{x}\right)$

**DerivativesÂ formulaÂ ofÂ Inverse Hyperbolic Functions**

\[\large \frac{d}{dx}sinh^{-1}x=\frac{1}{\sqrt{x^{2}+1}}\]

\[\large \frac{d}{dx}cosh^{-1}x=\frac{1}{\sqrt{x^{2}-1}}\]

\[\large \frac{d}{dx}tanh^{-1}x=\frac{1}{1-x^{2}}\]

\[\large \frac{d}{dx}coth^{-1}x=\frac{1}{1-x^{2}}\]

\[\large \frac{d}{dx}sech^{-1}x=\frac{-1}{x\sqrt{1-x^{2}}}\]

\[\large \frac{d}{dx}csch^{-1}x=\frac{-1}{|x|\sqrt{1+x^{2}}}\]