Math Doubts

Derivative Rule of Inverse Hyperbolic Cotangent function


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


Let $x$ represents a variable, the inverse hyperbolic co-tangent function is written in two different forms $\coth^{-1}{(x)}$ or $\operatorname{arccoth}{(x)}$ in mathematics. Now, the differentiation of the inverse hyperbolic cot function with respect to $x$ is written in the following mathematical forms.

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

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

The derivative of inverse hyperbolic cotangent function is also written as $(\coth^{-1}{x})’$ or $(\operatorname{arccoth}{x})’$ simply in differential calculus.

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

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

Other forms

The derivative of inverse hyperbolic cotangent function can be written in any variable in mathematics.


$(1) \,\,\,$ $\dfrac{d}{dt}{\,\coth^{-1}{t}}$ $\,=\,$ $\dfrac{1}{1-t^2}$

$(2) \,\,\,$ $\dfrac{d}{dv}{\,\coth^{-1}{v}}$ $\,=\,$ $\dfrac{1}{1-v^2}$

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


Learn how to derive the derivative formula for inverse hyperbolic cotangent 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