# Derivative of Inverse Hyperbolic Sine

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

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

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

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

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

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

Example: Find the derivative of $y = f\left( x \right) = {\sinh ^{ – 1}}4x$

We have the given function as
$y = {\sinh ^{ – 1}}4x$

Differentiating with respect to variable $x$, we get
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\sinh ^{ – 1}}4x$

Using the rule, $\frac{d}{{dx}}\left( {{{\sinh }^{ – 1}}x} \right) = \frac{1}{{\sqrt {1 + {x^2}} }}$, we get
$\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 + {{\left( {4x} \right)}^2}} }}\frac{d}{{dx}}4x \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{4}{{\sqrt {1 + 16{x^2}} }} \\ \end{gathered}$