There's a whole lot of chain rule going on. Let's see what you get.
I get derivative step by step, beginning by :
Tanª = 1/ (sec²(v))
{Here I use, v = (sec√x) and a = -1}
Then,
Ln'(v) = 1/v · v'
Next, internal derivative of v :
(sec√x · Tan√x) · ½ · xⁿ
{I use n = - ½ }
So, my entire equation would be :
g(x)' = 1/ (sec²(v) · v'/v · (sec √x · Tan√x)/2 · xⁿ)
{I use a = -½}
Then,
g'(x)' = √x / (sec²(Ln [sec√x]) · sec√x · Tan√x)
But I doubt that is correct and I don't know if is possible more simplification
(I'm sorry for the traslation the terms)