## Calculus: Early Transcendentals 8th Edition

$y'=2(tan^{-1}(x))(\frac{1}{1+x^2})$
Take the derivative and apply the chain rule using the derivative for inverse tangent ($\frac{1}{1+x^2}$): $y'=2(tan^{-1}(x))(\frac{1}{1+x^2})$