Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 34

$$y'=\frac{1}{1-x^2}$$

$$y=\frac{1}{2}\ln\frac{1+x}{1-x}$$ We have $$y'=\frac{1}{2}\frac{1}{\frac{1+x}{1-x}}\Big(\frac{1+x}{1-x}\Big)'=\frac{1}{2}\frac{1-x}{1+x}\frac{(1+x)'(1-x)-(1+x)(1-x)'}{(1-x)^2}$$ $$y'=\frac{1-x}{2(1+x)}\times\frac{1-x-(1+x)(-1)}{(1-x)^2}$$ $$y'=\frac{1-x}{2(1+x)}\times\frac{1-x+1+x}{(1-x)^2}$$ $$y'=\frac{1-x}{2(1+x)}\times\frac{2}{(1-x)^2}$$ $$y'=\frac{1}{(1+x)(1-x)}$$ $$y'=\frac{1}{1-x^2}$$