#### Answer

$$\frac{{ds}}{{dt}} = \frac{3}{{{{\left( {15t - 1} \right)}^4}}}$$

#### Work Step by Step

$$\eqalign{
& s = - \frac{1}{{15{{\left( {15t - 1} \right)}^3}}} \cr
& {\text{rewrite the function}} \cr
& s = - \frac{1}{{15}}{\left( {15t - 1} \right)^{ - 3}} \cr
& {\text{differentiate with respect to }}t \cr
& \frac{{ds}}{{dt}} = - \frac{1}{{15}}\frac{d}{{dt}}\left[ {{{\left( {15t - 1} \right)}^{ - 3}}} \right] \cr
& {\text{use the chain rule}} \cr
& \frac{{ds}}{{dt}} = - \frac{1}{{15}}\left( { - 3} \right){\left( {15t - 1} \right)^{ - 4}}\frac{d}{{dt}}\left[ {15t - 1} \right] \cr
& {\text{find the derivative and simplify}} \cr
& \frac{{ds}}{{dt}} = \frac{1}{5}{\left( {15t - 1} \right)^{ - 4}}\left( {15} \right) \cr
& \frac{{ds}}{{dt}} = 3{\left( {15t - 1} \right)^{ - 4}} \cr
& \frac{{ds}}{{dt}} = \frac{3}{{{{\left( {15t - 1} \right)}^4}}} \cr} $$