Answer
$ -\displaystyle \frac{4}{3x^{2}}+\frac{0.2}{x^{1.1}}+\frac{1.1x^{0.1}}{3.2}$
Work Step by Step
Written in exponential form,
... sum/difference rule...
$f^{\prime}(x)=\displaystyle \frac{d}{dx}[\frac{4}{3}x^{-1}]-\frac{d}{dx}[2x^{-0.1}]+\frac{d}{dx}[\frac{1}{3.2}x^{1.1}]-\frac{d}{dx}[4]$
... constant multiple ...
$=\displaystyle \frac{4}{3}\frac{d}{dx}[x^{-1}]-2\frac{d}{dx}[x^{-0.1}]+\frac{1}{3.2}\frac{d}{dx}[x^{1.1}]-\frac{d}{dx}[4]$
... power rule...
$=\displaystyle \frac{4}{3}(-x^{-2})-2(-0.1x^{-1.1})+\frac{1}{3.2}(1.1x^{0.1})-0$
$=-\displaystyle \frac{4}{3x^{2}}+\frac{0.2}{x^{1.1}}+\frac{1.1x^{0.1}}{3.2}$
