## Calculus (3rd Edition)

$$y'=\tan^{-1} x+\frac{x}{1+x^2}.$$
Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\tan^{-1} x)'=\dfrac{1}{1+x^2}$ Since $y=x\tan^{-1} x$, the derivative $y'$ by using the product rule, is given by $$y'=\tan^{-1} x+\frac{x}{1+x^2}.$$