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.

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