Find `(dy)/(dx)` in the following :
`y = tan^(-1)((3x-x^(3))/(1-3x^(2))), -(1)/(sqrt(3)) lt x lt (1)/(sqrt(3))`.

Explanation

`tan^(-1) ((3a^(2) x- x^(3))/(a^(3) - 3ax^(2)))`
Put `x = a tan theta " or " (x)/(a) = tan theta " or " theta = tan^(-1).(x)/(a)`
We have
`". tan^(-1) ((3a^(2) x -x^(3))/(a^(3) - 3ax^(2))) = tan^(-1) ((3a^(2). a tan theta - a^(3) tan^(3) theta)/(a^(3) - 3a. a^(2) tan^(2) theta))`
`= tan^(-1) ((3a^(3) tan theta - a^(3) tan^(3) theta)/(a^(3) - 3a^(3) tan^(2) theta))`
`= tan^(-1) ((3 tan theta - tan^(3) theta)/(1- 3 tan^(2) theta))`
`= tan^(-1) (tan 3 theta) = 3 theta`
`= 3 tan^(-1).(x)/(a)`