How do you find the derivative of inverse trig functions `y= arctan(x^2-1)^(1/2) + arc csc(x)` when x>1?

Assuming the equation was meant to be read as

`y = arctan(sqrt(x^2-1)) + arc csc(x)`

We can say the derivative will be the sum of the two other derivatives

`u = arctan(sqrt(x^2-1))`
`v = arc csc(x)`

`v` will be easier to differentiate, know, knowing that

`csc(arc csc(x)) = x`

We differentiate both sides and use the chain rule, so

`1 = csc^'(arc csc(x))*arc csc^'(x)`
`1/csc^'(arc csc(x)) = arc csc^'(x)`

We know that the derivative of the cossecant is `-csc(x)cot(x)`, so, in other words

`1/(-csc(arc csc(x))cot(arc csc(x))) = arc csc^'(x)`

Since they're both rational functions we can bring them up to the numerator

`-tan(arc csc(x))/x = arc csc^'(x)`

Using the pythagorean identity `1 + cot^2(theta) = csc^2(theta)` we have

`1 + 1/(tan^2(arc csc(x))) = x^2`

`1/(tan^2(arc csc(x))) = x^2 - 1`

`tan(arc csc(x)) = 1/sqrt(x^2 - 1)` so

`arc csc^'(x) = -1/(xsqrt(x^2-1))`

(There are authors that say it like this, there are people that have that leading x as `|x|`, considering we're working with `x > 1`, the absolute value bars aren't needed regardless of your philosophy if they should be there or not).

Now, we do `u = arctan(sqrt(x^2-1))`, for the sake of brevity, I'll skip the proof of the arctan's derivative, as the process is much like the one used for the arccsc, except it'd involve the `tan^2(theta) + 1 = sec^2(theta)` identity.

So we have

`u = arctan(sqrt(x^2-1))`
`u^' = arctan^'(sqrt(x^2-1))*sqrt((x^2-1))^'*2x`

It's important to remember we need to apply the chain rule twice here.

`u^' = 1/(1+(sqrt(x^2-1))^2)*1/(2sqrt(x^2-1))*2x`
`u^' = 1/(1+x^2-1)*1/(2sqrt(x^2-1))*2x`
`u^' = 1/x^2*1/(2sqrt(x^2-1))*2x`
`u^' = 1/(xsqrt(x^2-1))`

Summing them up we have

`y^' = 1/(xsqrt(x^2-1)) - 1/(xsqrt(x^2-1)) = 0`

Which makes sense if you look at the graph for `x > 0`. For smaller values the absolute value probably would have mattered and would have made the derivative non-zero.