Math Doubts

Derivative Rule of Inverse Hyperbolic Tangent function


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


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.


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

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

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


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

Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more