Chapter 3: Derivatives - Section 3.9 - Linearization and Differentials - Exercises 3.9 - Page 175: 42

$-tan^{-1}(\frac{x}{2})$

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