Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 35

$|sec x|$

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