## Thomas' Calculus 13th Edition

$\dfrac{(\sec x \tan x)}{|\tan x|}$
Given: $y=\cosh^{-1} (\sec x)$ Since, $\dfrac{d (\cosh^{-1} x)}{dx}=\dfrac{1}{ \sqrt{x^2 -1}}$ Apply chain rule to get the differentiation. Thus, $\dfrac{dy}{d x}=\dfrac{1}{ \sqrt{sec^2 x-1}}(\sec x \tan x)=\dfrac{1}{ \sqrt{ tan^2 x}}(\sec x \tan x)$ or, $\dfrac{dy}{d x}=\dfrac{(\sec x \tan x)}{|\tan x|}$