Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 74

Given,
y = tanh^-1(x)+ ln(sqrt(1-x^2))
so we have to find the y'
so,
y' =(tanh^-1(x)+ ln(sqrt(1-x^2)))'
=(tanh^-1(x))'+(ln(sqrt(1-x^2)))'
as we know
(tanh^-1(x))' =1/(1-x^2)
and so we have to find out
(ln(sqrt(1-x^2)))'
let u =sqrt( 1-x^2)
it can be solved by the following,
(df)/dx = (df)/(du) * (du)/dx
so,
d/dx (ln(sqrt(1-x^2)))= d/(du) ln(u) * (du)/dx
= 1/u * d/dx (sqrt( 1-x^2))
=1/sqrt( 1-x^2) * (-x/sqrt( 1-x^2))
= -x/( 1-x^2)
now,as
y'=(tanh^-1(x)+ ln(sqrt(1-x^2)))'
=1/( 1-x^2) +(-x/( 1-x^2))
= (1-x)/( 1-x^2)
=1/(1+x)