y=tan(2tan^-1x/2) show that dy/dx=4(1+y^2)/4+x^2
let u = tan^(-1)(x/2), with y = tan(2u)
du/dx = (1/2)*1/(1 + (x/2)^2)
du/dx = 2/(4 + x^2)
dy/du = 2sec^2(2u) = 2(1 + tan^2(2u))
dy/du = 2(1 + y^2)
Then,
dy/dx = (dy/du)(du/dx)
dy/dx = 2(1 + y^2)*2/(4 + x^2)
dy/dx = 4(1 + y^2)/(4 + x^2)