Question 248394
Let *[Tex \LARGE u=-2x] and *[Tex \LARGE v=x^2+5]. So *[Tex \LARGE u^{\prime}=-2] and *[Tex \LARGE v^{\prime}=2x].



*[Tex \LARGE \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu^{\prime}-uv^{\prime}}{v^2}] ... Start with the quotient rule


*[Tex \LARGE \frac{d}{dx}\left(\frac{-2x}{x^2+5}\right)=\frac{(x^2+5)(-2)-(-2x)(2x)}{(x^2+5)^2}] ... Plug in *[Tex \LARGE u=-2x, v=x^2+5, u^{\prime}=-2, \ \text{and} v^{\prime}=2x]



*[Tex \LARGE \frac{d}{dx}\left(\frac{-2x}{x^2+5}\right)=\frac{-2x^2-10+4x^2}{(x^2+5)^2}] ... Distribute



*[Tex \LARGE \frac{d}{dx}\left(\frac{-2x}{x^2+5}\right)=\frac{2x^2-10}{(x^2+5)^2}] ... Combine like terms.



So the derivative of {{{ -(2x)/(x^2+5) }}} is {{{(2x^2-10)/(x^2+5)^2}}}