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

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.

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

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.

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