## Calculus 10th Edition

$\frac{-2x}{(x^{2}+4)^{2}}$
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: $\frac{(x^{2}+4)\times (0)-(1)\times(2x)}{(x^{2}+4)^{2}}$ $=\frac{-2x}{(x^{2}+4)^{2}}$