Answer
$=-\displaystyle \frac{4}{x^{2}}+\frac{1}{4}-\frac{|x|}{x}$
Work Step by Step
Written in exponential form,
... sum/difference rule...
$f^{\prime}(x)=\displaystyle \frac{d}{dx}[4x^{-1}]+\frac{d}{dx}[\frac{1}{4}x]-\frac{d}{dx}[|x|]$
... constant multiple ...
$=4\displaystyle \frac{d}{dx}[x^{-1}]+\frac{1}{4}\frac{d}{dx}[x]-\frac{d}{dx}[|x|]$
... power rule, absolute value...
$=4(-x^{-2})+\displaystyle \frac{1}{4}(1)-\frac{|x|}{x}$
$=-\displaystyle \frac{4}{x^{2}}+\frac{1}{4}-\frac{|x|}{x}$
