Derivative of Inverse Hyperbolic Cotangent

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

Let the function be of the form \[y = f\left( x \right) = {\coth ^{ – 1}}x\]

By the definition of the 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
\[\begin{gathered} \frac{d}{{dx}}\coth y = \frac{d}{{dx}}\left( x \right) \\ \Rightarrow – {\operatorname{csch} ^2}y\frac{{dy}}{{dx}} = 1 \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{1}{{{{\operatorname{csch} }^2}y}}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered} \]

From the fundamental rules of inverse hyperbolic identities, this can be written as $${\operatorname{csch} ^2}y = {\coth ^2}y – 1$$. 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 $$\coth y = x$$, thus
\[\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{1 – {x^2}}} \\ \Rightarrow \frac{d}{{dx}}\left( {{{\coth }^{ – 1}}x} \right) = \frac{1}{{1 – {x^2}}} \\ \end{gathered} \]

Example: Find the derivative of \[y = f\left( x \right) = {\coth ^{ – 1}}2{x^3}\]

We have the given function as
\[y = {\coth ^{ – 1}}2{x^3}\]

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