#### Answer

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

#### Work Step by Step

Given that $y$ = $\ln{(x^2+4)}+x\tan^{-1}{\frac{x}{2}}$, on applying the product rule and chain rule of the derivative, we get:
$\frac{dy}{dx}$ = ${\frac{1}{x^2+4}\times\frac{d(x^2+4)}{dx}}-({tan^{-1}{\frac{x}{2}+x{\frac{d\tan^{-1}{\frac{x}{2}}}{dx}}}})$
or $\frac{dy}{dx}$ = ${\frac{2x}{x^2+4}-\tan^{-1}{\frac{x}{2}}-x{\frac{1}{1+{\frac{x^2}{4}}}{\frac{1}{2}}}}$
$\frac{dy}{dx}$ = ${\frac{2x}{x^2+4}-\tan^{-1}{\frac{x}{2}}-{\frac{2x}{x^2+4}}}$
The final answer is:$\frac{dy}{dx} = -\tan^{-1}{\frac{x}{2}}$