`y = tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))),find dy/dx.`

Explanation

Putt `x = cos 2 theta` so that `theta = (1)/(2) cos^(-1) x`
and `2 theta in [0, pi] " or " thteta in [0,(pi)/(2)]`
Then, we have
`L.H.S. = tan^(-1) ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x)))`
`= tan^(-1) ((sqrt(1 + cos 2 theta) - sqrt(1 - cos 2 theta))/(sqrt(1 + cos 2 theta) - sqrt(1 - cos 2 theta)))`
`= tan^(-1) ((sqrt(2 cos^(2) theta) - sqrt(2 sin^(2) theta))/(sqrt(2 cos^(2) theta) + sqrt(2 sin^(2) theta)))`
`= tan^(-1) ((sqrt2 cos theta - sqrt2 sin theta)/(sqrt2 cos theta + sqrt2 sin theta)) " " ( :' theta in [0, (pi)/(2)])`
`= tan^(-1) ((cos theta - sin theta)/(cos theta + sin theta))`
`= tan^(-1) ((1 - tan theta)/(1 + tan theta))`
`= tan^(-1) [tan ((pi)/(4) - theta)] " " ( :' (pi)/(4) - theta in [(-pi)/(4), (pi)/(4)])`
`= (pi)/(4) - theta = (pi)/(4) - (1)/(2) cos^(-1) x = R.H.S.`