Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 46

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

Given $$y=x\tanh^{-1}x$$ Then \begin{align*} y'&=\frac{d}{dx}\left(x\right)\tanh^{-1} \left(x\right)+\frac{d}{dx}\left(\tanh^{-1} \left(x\right)\right)x\\ &= \tanh^{-1} \left(x\right)+\frac{x}{1-x^2} \end{align*}