Differentiate `tan^(-1)((3x-x^3)/(1-3x^2))` , if `x<-1/(sqrt(3))`

Explanation

`f(x) = tan^-1 [ (3x-x^3)/(1-3x^2)]`
let `x= tan theta`
`theta= tan^-1 x`
so `f(x) = tan^-1 [ tan 3 theta] `
`x < -1/sqrt3`
`x= tan theta < -1/sqrt3`
`- pi/2 < theta < - pi/6`
`- (3 pi)/2 < 3 theta < -pi/2`
`f(x) = tan^-1(tan(3 theta))`
`= tan^-1 (tan(pi + 3 theta))`
`= pi + 3 theta`
`f(x) = pi + 3 tan^-1 x`
`f'(x) = 3/(1+x^2)`
Answer