# Derivative of Inverse Hyperbolic Secant

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

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

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

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

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

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

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

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

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

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