Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 27

$$y'=\tan^{-1} \left(4x\right)+\frac{4x}{16x^2+1}$$

Given $$y=x\tan^{-1}\left(4x\right)$$ Then \begin{align*} y'&=\frac{d}{dx}\left(x\right)\tan^{-1} \left(4x\right)+\frac{d}{dx}\left(\tan^{-1} \left(4x\right)\right)x\\ &=\tan^{-1} \left(4x\right)+\frac{4x}{16x^2+1} \end{align*}