Question

To find the derivative of the function.

Short answer

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

Step-by-step solution

 Given function

The function is \(y = {\coth ^{ - 1}}(\sec x)\).

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 cot function\(\frac{d}{{dx}}\left( {cot{h^{ - 1}}x} \right) = \frac{1}{{1 - {x^2}}}\).

Evaluate the derivative by the use of concepts

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

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

Let \(u = \sec x\) and apply the chain rule.

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

Substitute\(u = \sec x\).

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

Therefore, \(\frac{d}{{dx}}(\sec x) = \sec x\tan x\) and \(1 + {\tan ^2}x = {\sec ^2}x\).

\(\begin{array}{c} = \frac{{\sec x}}{{ - {\mathop{\rm lan}\nolimits} x}}\\ = \frac{{ - \frac{1}{{\sin x}}}}{{\frac{{\ln x}}{{\cot x}}}}\\\frac{{dy}}{{dx}} = - \frac{1}{{\cos x}} \times \frac{{\cos x}}{{\sin x}}\\ = - \frac{1}{{\sin x}}\end{array}\)

So, \( = - \csc x\).

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