Derivative of Inverse Hyperbolic Cotangent

In this tutorial we shall be concerned with the derivative of inverse hyperbolic tangent function with an example.

Let the function of the form

y = f\left( x \right) = {\coth ^{ - 1}}x

By definition of inverse trigonometric function, y = {\coth ^{ - 1}}x can be written as

\coth y = x

Differentiating both sides with respect to the variable x, we have

\begin{gathered} \frac{d}{{dx}}\coth y = \frac{d}{{dx}}\left( x \right) \\ \Rightarrow - {\operatorname{csch} ^2}y\frac{{dy}}{{dx}} = 1 \\ \Rightarrow \frac{{dy}}{{dx}} = - \frac{1}{{{{\operatorname{csch} }^2}y}}\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right) \\ \end{gathered}

We can write from the fundamental rules of inverse hyperbolic identities {\operatorname{csch} ^2}y = {\coth ^2}y - 1. Putting this value in above relation (i) and simplifying, we have

\begin{gathered} \frac{{dy}}{{dx}} = - \frac{1}{{{{\coth }^2}y - 1}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{1 - {{\coth }^2}y}} \\ \end{gathered}

From the above we have \coth y = x, by putting this

\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{1 - {x^2}}} \\ \Rightarrow \frac{d}{{dx}}\left( {{{\coth }^{ - 1}}x} \right) = \frac{1}{{1 - {x^2}}} \\ \end{gathered}

Example: Find the derivative of

y = f\left( x \right) = {\coth ^{ - 1}}2{x^3}

We have the given function as

y = {\coth ^{ - 1}}2{x^3}

Differentiation with respect to variable x, we get

\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\coth ^{ - 1}}2{x^3}

Using the rule, \frac{d}{{dx}}\left( {{{\coth }^{ - 1}}x} \right) = \frac{1}{{1 - {x^2}}}, we get

\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{1 - {{\left( {2{x^3}} \right)}^2}}}\frac{d}{{dx}}2{x^3} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{1 - 2{x^6}}}\left( {6{x^2}} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{6{x^2}}}{{1 - 2{x^6}}} \\ \end{gathered}