Before answering let us differentiate the inverse tan function.
#arctan(x)=y#
#tan(arctan(x))=tan(y)#
#d/dx[x]=d/dx[tan(y)]#
#1=sec^2(y)*(dy)/(dx)#
#1/(sec^2(arctanx))=(dy) /(dx) #
But #sqrt(tan^2(alpha)+1)=sec^2(alpha)#
Then in which #alpha=arctan(x)#:
#1/(sqrt(tan^2(arctan(x))+1))=(dy) /(dx) #
#1/(sqrt(x^2+1))=(dy)/(dx) #
Now differentiating:
#d/dx[arctan(x-(1+x^2)^(1/2))]#
#=[1/sqrt((x-(1+x^2)^(1/2))^2+1)]d/dx[x-(1+x^2)^(1/2)]#
#=[1/sqrt((x-(1+x^2)^(1/2))^2+1)][1-(1/2)(1+x^2)^(-1/2)d/dx[x^2+1]]#
#=[1/sqrt((x-(1+x^2)^(1/2))^2+1)][1-((1+x^2)^(-1/2))/2[2x]]#
#=[1/sqrt((x-(1+x^2)^(1/2))^2+1)][1-x(1+x^2)^(-1/2)]#
