Answer
$\displaystyle \frac{dy}{dx}=\frac{|x| }{x^{3}}$
Work Step by Step
(At the end of section 10-6, we found: $\displaystyle \frac{d}{dx}[\ |x|\ ]=\frac{|x|}{x}$ )
$f(x)=x^{2},\displaystyle \ \ \quad g(x)=|x|,\quad y=\frac{f(x)}{g(x)}$
$f^{\prime}(x)=2x\displaystyle \qquad g^{\prime}(x)=\frac{|x|}{x}$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[\frac{f(x)}{g(x)}]$= ... quotient rule ...
$= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$=\displaystyle \frac{2x\cdot|x| -x^{2}\cdot\frac{|x|}{x}}{(|x|)^{2}}\qquad ... |x|=x^{2}$
$=\displaystyle \frac{ 2x|x|-x|x| }{x^{2}}=$
$=\displaystyle \frac{x|x| }{x^{2}}$
$\displaystyle \frac{dy}{dx}=\frac{|x| }{x}$