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