Question

To find the derivative of the function.

Short answer

The derivative of the function \(y = {\sinh ^{ - 1}}(\tan x)\) is, \(\frac{{dy}}{{dx}} = |\sec x|\).

Step-by-step solution

Given function

The function is \(y = {\sinh ^{ - 1}}(\tan x)\).

The Concept of derivative and hyperbolic function

Derivative rules:

(1) The chain rule:\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\)

(2) The derivative of the inverse hyperbolic sine function is,\(\frac{d}{{dx}}\left( {sin{h^{ - 1}}x} \right) = \frac{1}{{\sqrt {1 + {x^2}} }}\).

Explanation of Solution by the concepts above

The derivative of the function \(y = {\sinh ^{ - 1}}(\tan x)\) is computed as shown below.

\(\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{{\sinh }^{ - 1}}(\tan x)} \right)\)

Let \(u = \tan x.\)

\(\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{{\sinh }^{ - 1}}u} \right)\)

Apply the chain rule and simplify the terms.

\(\begin{array}{c}\frac{{dy}}{{dx}} = \frac{d}{{du}}\left( {{{\sinh }^{ - 1}}u} \right)\frac{{du}}{{dx}}\\ = \frac{1}{{\sqrt {1 + {u^2}} }}\frac{{du}}{{dx}}\end{array}\)

Substitute\(u = \tan x\).

\(\begin{array}{c}\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 + {{(\tan x)}^2}} }}\frac{d}{{dx}}(\tan x)\\ = \frac{1}{{\sqrt {1 + {{\tan }^2}x} }}\left( {{{\sec }^2}x} \right)\\ = \frac{{{{\sec }^2}x}}{{\sqrt {{{\sec }^2}x} }}\left( {1 + {{\sec }^2}x = {{\tan }^2}x} \right)\\ = \frac{{{{\sec }^2}x}}{{|\sec x|}}\end{array}\)

Simplify the expression as follows:

\(\begin{array}{r}\frac{{dy}}{{dx}} = |\sec x|\quad \\\left( {|\sec x| = \sqrt {{{\sec }^2}x} } \right)\end{array}\)

Therefore, the derivative of the function \(y = {\sinh ^{ - 1}}(\tan x)\) is, \(\frac{{dy}}{{dx}} = |\sec x|\).