Answer
$y'=-\csc x$
Work Step by Step
Given: $y=\coth^{-1}(\sec x)$
Now, $y'=\dfrac{d}{dx}[\coth^{-1}(\sec x)]$
Apply chain rule, we have
$y'=\dfrac{d\coth^{-1}(\sec x)}{d\sec x}( \dfrac{d\sec x}{dx})$
or, $y'=\dfrac{1}{1-\sec^2 x}(\sec x\tan x)$
Use formula: $\sec^2 x-1=\tan^2 x$
,
Therefore, $y'=\dfrac{1}{-\tan^2 x}(\sec x\tan x)=\dfrac{-\sec x}{\tan x}$
or, $=\frac{-1}{\sin x}$
Thus,$y'=-\csc x$
