differentiate tan-1(2x/1-x^2)

you know that

d/du tan-1(u) = 1/(1+u^2)

So, use the chain rule and you have

u = 2x/(1-x^2)
du/dx = 2(1+x^2)/(1-x^2)^2
so, the derivative is

1/(1+(2x/(1-x^2))^2) * 2(1+x^2)/(1-x^2)^2 = 2/(1+x^2)

to see why this simplifies no neatly, consider that if

x = tanθ
2x/(1-x^2) = tan2θ

arctan(2x/(1-x^2)) = arctan(tan(2θ)) = 2θ = 2arctan(x)

the derivative is now 2/(1+x^2)