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.
