Answer
$$y'=|\sec x|$$
Work Step by Step
$y'=\frac{d}{dx}\sinh^{-1}(\tan x)$
Using the chain rule:
$y'=\frac{d\sinh^{-1}(\tan x)}{d\tan x}
\times \frac{d\tan x}{dx}$
$=\frac{1}{\sqrt{1+(\tan x)^2}}
\times \sec^2 x$
Recall that:
$$\tan^2 x+1=\sec^2 x$$
Thus,
$y'=\frac{1}{\sqrt{1+\tan^2 x}}
\times \sec^2 x$
$=\frac{1}{|\sec x|}\times \sec^2 x$
$=|\sec x|$
