How do you find the derivative of #y= arctan sqrt(x^2 -1) + arc csc x#?

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Answer 1 · 2017-01-07T02:25:42

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Use # f ^(-1)f(a)=a#.

To make y real x in [-1, 1]. Befittingly,

let #x = sec theta in [-1, 1]#.

Now,

#y = arc tan (tan theta) + pi/2 - arc sec (sec theta)#,

using #arc csc x+arc sec x = pi/2#

#=theta +pi/2-theta#

#=pi/2# And so,

#y' = 0#.