Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 32

$$ y'= \frac{1}{(1-x^2)\tanh^{-1}x}.$$

Since $ y=\ln (\tanh^{-1}x)$, then the derivative, by using the chain rule, is given by $$ y'= \frac{1}{\tanh^{-1}x}(\tanh^{-1}x )'=\frac{1}{\tanh^{-1}x}\frac{1 }{1-x^2}=\frac{1}{(1-x^2)\tanh^{-1}x}.$$