# Derivative of Inverse Hyperbolic Cosecant

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

Let the function be of the form $y = f\left( x \right) = {\operatorname{csch} ^{ – 1}}x$

By the definition of the inverse trigonometric function, $y = {\operatorname{csch} ^{ – 1}}x$ can be written as
$\operatorname{csch} y = x$

Differentiating both sides with respect to the variable $x$, we have
$\begin{gathered} \frac{d}{{dx}}\operatorname{csch} y = \frac{d}{{dx}}\left( x \right) \\ \Rightarrow – \csc {\text{h}}y\coth y\frac{{dy}}{{dx}} = 1 \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{1}{{\csc {\text{h}}y\coth y}}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered}$

From the fundamental rules of inverse hyperbolic identities, this can be written as $\coth y = \sqrt {1 + \csc {{\text{h}}^2}x}$. Putting this value in above relation (i) and simplifying, we have
$\frac{{dy}}{{dx}} = – \frac{1}{{\csc {\text{h}}x\sqrt {1 + \csc {{\text{h}}^2}x} }}$

From the above we have $\operatorname{csch} y = x$, thus
$\begin{gathered} \frac{{dy}}{{dx}} = – \frac{1}{{x\sqrt {1 + {x^2}} }} \\ \Rightarrow \frac{d}{{dx}}\left( {{{\operatorname{csch} }^{ – 1}}x} \right) = – \frac{1}{{x\sqrt {1 + {x^2}} }} \\ \end{gathered}$

Example: Find the derivative of $y = f\left( x \right) = {\operatorname{csch} ^{ – 1}}2x$

We have the given function as
$y = {\operatorname{csch} ^{ – 1}}2x$

Differentiating with respect to variable $x$, we get
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\operatorname{csch} ^{ – 1}}2x$

Using the rule, $\frac{d}{{dx}}\left( {{{\operatorname{csch} }^{ – 1}}x} \right) = – \frac{1}{{x\sqrt {1 + {x^2}} }}$, we get
$\begin{gathered} \frac{{dy}}{{dx}} = – \frac{1}{{2x\sqrt {1 – {{\left( {2x} \right)}^2}} }}\frac{d}{{dx}}2x \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{1}{{2x\sqrt {1 – 4{x^2}} }}2\left( 1 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{1}{{x\sqrt {1 – 4{x^2}} }} \\ \end{gathered}$