Answer
$r^{\prime}(x)=\displaystyle \frac{8x}{3}-\frac{3.2x^{2.2}}{6}+\frac{4}{3x^{3}}$
Work Step by Step
SUMMARY:
The Power Rule$:\ \ \ [x^{n}]^{\prime}=nx^{n-1 } $
Sum Rule: $\ \ \ \ \ \ [f\pm g]^{\prime}(x)=f^{\prime}(x)\pm g^{\prime}(x) $
Constant Multiple Rule:$\ \ \ [cf]^{\prime}(x)=cf^{\prime}(x) $
Constant times x:$\ \ \ \displaystyle \frac{d}{dx}(cx)=c $
Constant:$\displaystyle \ \ \ \ \ \frac{d}{dx}(c)=0 $
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$r^{\prime}(x)=[ \displaystyle \frac{4}{3}x^{2} +\displaystyle \frac{1}{6}x^{3.2}-\frac{2}{3}x^{-2}+4]^{\prime}=... $Sum Rule,
$=[ \displaystyle \frac{4}{3}x^{2}]^{\prime} + [\displaystyle \frac{1}{6}x^{3.2}]^{\prime}-[\frac{2}{3}x^{-2}]^{\prime}+[4]^{\prime}=...$Constant Multiple Rule
$=\displaystyle \frac{4}{3}[ x^{2}]^{\prime} -\displaystyle \frac{1}{6}[x^{3.2}]^{\prime}-\frac{2}{3}[x^{-2}]^{\prime}+4[x^{0}]^{\prime}=$...Power Rule...
$=\displaystyle \frac{4}{3}[ 2x] -\displaystyle \frac{1}{6}[3.2x^{2.2}]-\frac{2}{3}[-2x^{-3}]^{\prime}+4[0]=$
$r^{\prime}(x)=\displaystyle \frac{8x}{3}-\frac{3.2x^{2.2}}{6}+\frac{4}{3x^{3}}$
