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