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,

$y' = 0$ .

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