differentiate tan^-1(1+ax/1-ax) with respect to sqrt(1+a^2x^2)

Hi

Let x = (1/a) tan yThen, u = tan^1 [(1 + tan y)/(1 - tan y)] = tan^1 [(tan pi/4 + tan y)/(1 - tan pi/4 tan y)] = tan^1 [tan (pi/4 +y)] = (pi/4 + y) = pi/4 + tan^1(ax)du/dx = 1/(1+a^2x^2)v = (1 + a^2x^2)^1/2dv/dx = a^2x/ ( (1 + a^2x^2)^1/2du/du = 1/a^2x (1 + a^2x^2)^1/2

Let x = (1/a) tan yThen, u = tan^-1 [(1 + tan y)/(1 - tan y)] = tan^-1 [(tan pi/4 + tan y)/(1 - tan pi/4 tan y)] = tan^-1 [tan (pi/4 +y)] = (pi/4 + y) = pi/4 + tan^-1(ax)du/dx = 1/(1+a^2x^2)v = (1 + a^2x^2)^1/2dv/dx = a^2x/ ( (1 + a^2x^2)^1/2du/du = 1/a^2x (1 + a^2x^2)^1/2