Answer
$$\frac{{dy}}{{dx}} = \frac{{ - 6{x^3} + 30{x^2} - 20x - 31}}{{{{\left( {{x^3} - 10{x^2} + 29x - 20} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{3x - 1}}{{\left( {x - 5} \right)\left( {x - 4} \right)\left( {x - 1} \right)}} \cr
& {\text{Simplify the denominator}} \cr
& y = \frac{{3x - 1}}{{\left( {{x^2} - 9x + 20} \right)\left( {x - 1} \right)}} \cr
& y = \frac{{3x - 1}}{{{x^3} - {x^2} - 9{x^2} + 9x + 20x - 20}} \cr
& y = \frac{{3x - 1}}{{{x^3} - 10{x^2} + 29x - 20}} \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{3x - 1}}{{{x^3} - 10{x^2} + 29x - 20}}} \right] \cr
& {\text{By the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {{x^3} - 10{x^2} + 29x - 20} \right)\left( 3 \right) - \left( {3x - 1} \right)\left( {3{x^2} - 20x + 29} \right)}}{{{{\left( {{x^3} - 10{x^2} + 29x - 20} \right)}^2}}} \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{3{x^3} - 30{x^2} + 87x - 60 - 9{x^3} + 60{x^2} - 87x - 20x + 29}}{{{{\left( {{x^3} - 10{x^2} + 29x - 20} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - 6{x^3} + 30{x^2} - 20x - 31}}{{{{\left( {{x^3} - 10{x^2} + 29x - 20} \right)}^2}}} \cr} $$
