## Thomas' Calculus 13th Edition

$-tan^{-1}(\frac{x}{2})$
$y=\ln (x^2+4)-x \tan^{-1}(\frac{x}{2})$ Derive using the chain rule: $\frac{dy}{dx}=\frac{2x}{x^2+4}-tan^{-1}(\frac{x}{2})-x[\frac{\frac{1}{2}}{1+(\frac{x}{2})^2}]$ $=\frac{2x}{x^2+4}-tan^{-1}(\frac{x}{2})-\frac{2x}{4+x^2}$ $=-tan^{-1}(\frac{x}{2})$