## Calculus 8th Edition

$y' =\frac{5}{3} x^{2/3}- \frac{2}{3x^{1/3}}$
$y=x^{5/3}-x^{2/3}$ Lets use the formula $(f-g)'=f' - g'$ then $y' = (x^{5/3}-x^{2/3})' = (x^{5/3})'-(x^{2/3})'$ by the formulas $(x^{n})'=nx^{n-1}$, $(c)' = 0$ lets choose $n=5/3$ So, $(x^{5/3})' = nx^{n-1} = \frac{5}{3} x^{5/3-1}=\frac{5}{3} x^{2/3}$ lets choose $n=2/3$ So, $(x^{2/3})' = nx^{n-1} = \frac{2}{3} x^{2/3-1}=\frac{2}{3} x^{-1/3}=\frac{2}{3} \frac{1}{x^{1/3}}$ therefore $y' = (x^{5/3}-x^{2/3})' = (x^{5/3})'-(x^{2/3})'=\frac{5}{3} x^{2/3}-\frac{2}{3x^{1/3}}$