Answer
$\displaystyle \frac{dy}{dx}=0$
Work Step by Step
(At the end of section 10-6, we found: $\displaystyle \frac{d}{dx}[\ |x|\ ]=\frac{|x|}{x}$ )
$f(x)=x,\displaystyle \ \ \quad g(x)=|x|,\quad y=\frac{f(x)}{g(x)}$
$f^{\prime}(x)=1\displaystyle \qquad g^{\prime}(x)=\frac{|x|}{x}$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[\frac{f(x)}{g(x)}]$= ... quotient rule ...
$= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$=\displaystyle \frac{1\cdot|x| - x\cdot\frac{|x|}{x}}{|x|^{2}}\qquad$....$|x|^{2}=x^{2}$
$=\displaystyle \frac{|x| - |x|}{x^{2}}=0$
