## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{(\sec x \tan x)}{|\tan x|}$
We are given: $y=\cosh^{-1} (\sec x)$ Recall the formula: $\dfrac{d (\cosh^{-1} x)}{dx}=\dfrac{1}{ \sqrt{x^2 -1}}$ We need to use the chain rule to get the differentiation: Thus, $\dfrac{dy}{d x}=\dfrac{1}{ \sqrt{sec^2 x-1}}(\sec x \tan x)=\dfrac{1}{ \sqrt{ tan^2 x}}(\sec x \tan x) =\dfrac{(\sec x \tan x)}{|\tan x|}$