Math Doubts

Derivative Rule of Inverse Hyperbolic Tangent function

Formula

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

Introduction

The inverse hyperbolic tangent is written in function form as $\tanh^{-1}{(x)}$ or $\operatorname{arctanh}{(x)}$ if the literal $x$ represents a variable. The differentiation of the inverse hyperbolic tan function with respect to $x$ is written in the following mathematical forms.

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

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

In simple form, the derivative of inverse hyperbolic tan function is written as $(\tanh^{-1}{x})’$ or $(\operatorname{arctanh}{x})’$ mathematically in differential calculus.

The differentiation of hyperbolic inverse tangent function with respect to $x$ is equal to multiplicative inverse of difference of $x$ squared from one.

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

Other forms

The derivative of inverse hyperbolic tangent function can also be expressed in any variable in mathematics.

Example

$(1) \,\,\,$ $\dfrac{d}{dl}{\, \tanh^{-1}{l}}$ $\,=\,$ $\dfrac{1}{1-l^2}$

$(2) \,\,\,$ $\dfrac{d}{dq}{\, \tanh^{-1}{q}}$ $\,=\,$ $\dfrac{1}{1-q^2}$

$(3) \,\,\,$ $\dfrac{d}{dy}{\, \tanh^{-1}{y}}$ $\,=\,$ $\dfrac{1}{1-y^2}$

Proof

Learn how to prove differentiation rule of inverse hyperbolic tangent function from the first principle of differentiation.

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved