I have asumed that the question as written has omitted brackets around the #(1/2)# exponents so that:
#y= Ln [( x^(1/2)) / ( 1+x^(1/2))]#
If this is the case then:
#dy/dx = (1+x^(1/2))/(x^(1/2)# * #d/dx (( x^(1/2)) / ( 1+x^(1/2)))# Standard differential and chain rule
#dy/dx = (1+x^(1/2))/(x^(1/2)# * # [(1+x^(1/2)) * 1/2x^(-1/2) - x^(1/2) * 1/2x^(-1/2)] / (1+x^(1/2))^2# Quotient rule
#= (1+x^(1/2))/(x^(1/2) # * # 1/2x^(-1/2) [(1+x^(1/2)-x^(1/2))]/ (1+x^(1/2))^2#
#= (1+x^(1/2))/(x^(1/2) # * # [(1+cancel(x^(1/2))-cancel(x^(1/2)))] / [2x^(1/2)(1+x^(1/2))^2]#
#=[ (1+x^(1/2))] / [2x (1+x^(1/2))^2]#
#= cancel((1+x^(1/2)))/ [2x (1+x^(1/2))^cancel(2])#
#= 1/[2(x+x^(3/2))]#
