Math Doubts

Derivative Rule of Inverse Hyperbolic Cosine function

Formula

$\dfrac{d}{dx}{\, \cosh^{-1}{x}}$ $\,=\,$ $\dfrac{1}{\sqrt{x^2-1}}$

Introduction

The inverse hyperbolic cosine function is expressed as $\cosh^{-1}{(x)}$ or $\operatorname{arccosh}{(x)}$ mathematically when the $x$ denotes a variable. The derivative of the inverse hyperbolic cosine function with respect to $x$ is expressed in the below mathematical forms.

$(1).\,\,\,$ $\dfrac{d}{dx}{\, (\cosh^{-1}{x})}$

$(2).\,\,\,$ $\dfrac{d}{dx}{\, (\operatorname{arccosh}{x})}$

In mathematics, the derivative of inverse hyperbolic cosine function is written as $(\cosh^{-1}{x})’$ or $(\operatorname{arccosh}{x})’$ simply in differential calculus.

The differentiation of hyperbolic inverse cos function with respect to $x$ is equal to reciprocal of the square root of difference of $1$ from $x$ squared.

$\implies$ $\dfrac{d}{dx}{\, \cosh^{-1}{x}}$ $\,=\,$ $\dfrac{1}{\sqrt{x^2-1}}$

Other forms

The derivative of inverse hyperbolic cosine function can also be written in terms of any variable in mathematics.

Example

$(1) \,\,\,$ $\dfrac{d}{de}{\, \cosh^{-1}{e}}$ $\,=\,$ $\dfrac{1}{\sqrt{e^2-1}}$

$(2) \,\,\,$ $\dfrac{d}{dm}{\, \cosh^{-1}{m}}$ $\,=\,$ $\dfrac{1}{\sqrt{m^2-1}}$

$(3) \,\,\,$ $\dfrac{d}{dz}{\, \cosh^{-1}{z}}$ $\,=\,$ $\dfrac{1}{\sqrt{z^2-1}}$

Proof

Learn how to prove differentiation rule of hyperbolic inverse cosine function by the first principle of differentiation.

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