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

Question

10230 views around the world

You can reuse this answer
Creative Commons License

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)#