How do you find the derivative of $y = cosh^-1 (secx)$ ?

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Answer 1 · 2016-07-23T13:58:20

$\ y' = (sec x tan x)/ |tan x|$

$= sec x sgn (tan x)$

re-qrite slightly

$cosh y = sec x$

then diff implicitly

$sinh y \ y' = sec x tan x$

$\ y' = (sec x tan x)/sinh y$

$\ y' = (sec x tan x)/ (sqrt( cosh^2 y - 1 ))$

$\ y' = (sec x tan x)/ (sqrt( sec^2 x - 1 ))$

$\ y' = (sec x tan x)/ (sqrt( tan^2 x ))$

$\ y' = (sec x tan x)/ |tan x|$

$= sec x sgn (tan x)$

How do you find the derivative of y = cosh^-1 (secx)y = cosh^-1 (secx)?