Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 35

$$H'(v) =\tan^{-1} \left(v\right)+\frac{v}{v^2+1}$$

Given $$H(v)=v\tan^{-1}v$$ Then \begin{align*} H'(v) &=\frac{d}{dv}\left(v\right)\tan^{-1} \left(v\right)+\frac{d}{dv}\left(\tan^{-1} \left(v\right)\right)v\\ &=\tan^{-1} \left(v\right)+\frac{v}{v^2+1} \end{align*}