Answer
$\displaystyle \frac{dy}{dx}=-\frac{|x| }{x^{3}}$
Work Step by Step
(At the end of section 10-6, we found: $\displaystyle \frac{d}{dx}[\ |x|\ ]=\frac{|x|}{x}$ )
$f(x)=|x|,\displaystyle \ \ \quad g(x)=x^{2},\quad y=\frac{f(x)}{g(x)}$
$f^{\prime}(x)=\displaystyle \frac{|x|}{x}\qquad g^{\prime}(x)=2x$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[\frac{f(x)}{g(x)}]$= ... quotient rule ...
$= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$=\displaystyle \frac{\frac{|x|}{x}\cdot x^{2} - |x|\cdot 2x}{(x^{2})^{2}}$
$=\displaystyle \frac{x|x| - 2x|x|}{x^{4}}=$
$=-\displaystyle \frac{x|x| }{x^{4}}$
$\displaystyle \frac{dy}{dx}=-\frac{|x| }{x^{3}}$
