Derivative of Inverse Hyperbolic Cosecant

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

Let the function of the form

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

By definition of 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}

We can write from the fundamental rules of inverse hyperbolic identities \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, by putting this

\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

Differentiation 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}