Answer
$f^{\prime}(x)=50x^{4}+2x^{3}-1$
Work Step by Step
... sum/difference...
$f^{\prime}(x)=\displaystyle \frac{d}{dx}[10x^{5}]+\frac{d}{dx}[\frac{1}{2}x^{4}]-\frac{d}{dx}[x]+\frac{d}{dx}[2]$
... constant multiple ...
$=10\displaystyle \frac{d}{dx}[x^{5}]+\frac{1}{2}\frac{d}{dx}[x^{4}]-\frac{d}{dx}[x]+\frac{d}{dx}[2]$
... power, constant ...
$=10(5x^{4})+\displaystyle \frac{1}{2}(4x^{3})-1+0$
$f^{\prime}(x)=50x^{4}+2x^{3}-1$
