Answer
$ \displaystyle \frac{-50}{x^{6}}-\frac{2}{x^{5}}+\frac{1}{x^{2}}$
Work Step by Step
Written in exponential form,
... sum/difference rule...
$f^{\prime}(x)=\displaystyle \frac{d}{dx}[10x^{-5}]+\frac{d}{dx}[\frac{1}{2}x^{-4}]-\frac{d}{dx}[x^{-1}]+\frac{d}{dx}[2]$
... constant multiple ...
$=10\displaystyle \frac{d}{dx}[x^{-5}]+\frac{1}{2}\frac{d}{dx}[x^{-4}]-\frac{d}{dx}[x^{-1}]+\frac{d}{dx}[2]$
... power, constant rules...
$=10(-5x^{-6})+\displaystyle \frac{1}{2}(-4x^{-5})-(-x^{-2})+0$
$=\displaystyle \frac{-50}{x^{6}}-\frac{2}{x^{5}}+\frac{1}{x^{2}}$
