Question

y=arctan(sqrt(9x^2-1))^2

dy/dx=?

Answer 1

since (√(9x^2-1))^2 = 9x^2-1,

I assume you mean arctan^2 √(9x^2-1)
let u = 9x^2-1
y = arctan^2 (√u)
y' = 2 arctan(√u) * 1/(u+1) * 1/(2√u) u'
= 2 arctan(√9x^2-1) * 1/(9x^2) * 1/(2√(9x^2-1)) * 18x
= 2 arctan(√9x^2-1) / [x √(9x^2-1)]