Answer
$f'\left( t \right) = \frac{{ - 10{t^3} - 5}}{{{{\left( {{t^3} - t - 1} \right)}^2}}}$
Work Step by Step
$$\eqalign{
& f\left( t \right) = \frac{{5t}}{{{t^3} - t - 1}} \cr
& {\text{Differentiating}} \cr
& f'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{5t}}{{{t^3} - t - 1}}} \right] \cr
& {\text{Differentiate by using the formula }}\left( {\frac{g}{h}} \right)' = \frac{{hg' - gh'}}{{{h^2}}},{\text{ }}h \ne 0 \cr
& {\text{Let }}g = 5t{\text{ and }}h = {t^3} - t - 1,{\text{ then}} \cr
& f'\left( t \right) = \frac{{\left( {{t^3} - t - 1} \right)\left( {5t} \right)' - \left( {5t} \right)\left( {{t^3} - t - 1} \right)'}}{{{{\left( {{t^3} - t - 1} \right)}^2}}} \cr
& {\text{Compute derivatives}} \cr
& f'\left( t \right) = \frac{{\left( {{t^3} - t - 1} \right)\left( 5 \right) - \left( {5t} \right)\left( {3{t^2} - 1} \right)}}{{{{\left( {{t^3} - t - 1} \right)}^2}}} \cr
& {\text{Multiply and simplify}} \cr
& f'\left( t \right) = \frac{{5{t^3} - 5t - 5 - 15{t^3} + 5t}}{{{{\left( {{t^3} - t - 1} \right)}^2}}} \cr
& f'\left( t \right) = \frac{{ - 10{t^3} - 5}}{{{{\left( {{t^3} - t - 1} \right)}^2}}} \cr} $$
