#### Answer

$F'(r)=\frac{-15}{r^{4}}$

#### Work Step by Step

$F(r)=\frac{5}{r^{3}}$
Lets use the formula $(\frac{f}{g})'=\frac{gf'-fg'}{g^{2}}$
then
$F'(r)=(\frac{5}{r^{3}})'=\frac{r^{3} \times (5)' - 5 \times (r^{3})'}{r^{6}}$
by the formulas $(x^{n})'=nx^{n-1}$, $(c)' = 0$
$(r^{3})' = 3r^{2}$
$(5)' = 0$
therefore
$F(r)=\frac{5}{r^{3}}=\frac{r^{3} \times (5)' - 5 \times (r^{3})'}{r^{6}}=\frac{0- 5 \times 3r^{2}}{r^{6}}=\frac{-15r^{2}}{r^{6}}=\frac{-15}{r^{4}}$