Answer
$\frac{-2x}{(x^{2}+4)^{2}}$
Work Step by Step
Let the equation be of the form $y=\frac{u}{v}$
Let $u=1$
$\frac{du}{dx}=0$
Let $v=x^{2}+4$
$\frac{dv}{dx}=2x$
The equation for the derivative is:
$\frac{v \times \frac{du}{dx} - u\times\frac{dv}{dx}}{v^{2}}$
Using the above values, it can be found that the derivative is equal to:
$\frac{(x^{2}+4)\times (0)-(1)\times(2x)}{(x^{2}+4)^{2}}$
$=\frac{-2x}{(x^{2}+4)^{2}}$