Differentiating #ln(1+x)#
Say: #p=ln(1+x)=lnu#
#(dp)/(du)=1/u=1/(1+x)#
#u=1+x#, #(du)/(dx)=1#
Therefore:
#(dp)/(dx)=1/(1+x)#
Differentiating #ln(1-x)#
Say:
#q=ln(1-x)=lnu#
#(dq)/(du)=1/u=1/(1-x)#
#u=1-x#, #(du)/(dx)=-1#
Therefore:
#(dq)/(dx)=-1/(1-x)#
Differentiating #y=1/sqrt(1-x^2)#
#y=1/sqrt(1-x^2)#
#lny=ln(1/sqrt(1-x^2))#
#lny=ln1-ln((1-x^2)^(1/2))#
#lny=-1/2ln(1-x^2)#
#lny=-1/2ln((1+x)(1-x))#
#lny=-1/2{ln(1+x)+ln(1-x)}#
#lny=-1/2ln(1+x)-1/2ln(1-x)#
#1/y*(dy)/(dx)=-1/2{1/(1+x)}-1/2{-1/(1-x)}#
#1/y*(dy)/(dx)=1/(2(1-x))-1/(2(1+x))#
#y*1/y*(dy)/(dx)=y{1/(2(1-x))-1/(2(1+x))}#
#(dy)/(dx)=1/(sqrt(1-x^2)){1/(2(1-x))-1/(2(1+x))}#
#(dy)/(dx)=1/(sqrt((1+x)(1-x))){1/(2(1-x))-1/(2(1+x))}#
An explanation provided by Harivogind S. as to why the result above is the same as the result he has provided...
#{1/(2(1-x))-1/(2(1+x))} = [(1+x) - (1-x)]/[ 2*(1-x)*(1+x)]# - common denominator.
#= (2x)/[2*(1-x^2)]#
#{1/(2(1-x))-1/(2(1+x))} = x/(1-x^2)#
and,
#1/sqrt[(1+x)(1-x)] = 1/sqrt[(1-x^2)]#
Therefore -
#(dy)/(dx)=1/(sqrt((1+x)(1-x))){1/(2(1-x))-1/(2(1+x))} = 1/sqrt(1-x^2)*[x/(1-x^2)]#
#(dy)/(dx)=x/(1-x^2)^(3/2)#
*Quotient rule was not required in these workings as your fraction didn't contain both a numerator and denominator that were both functions of x.
