Answer
$F'\left( x \right) = \frac{{12x - 6{x^2}}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}}$
Work Step by Step
$$\eqalign{
& F\left( x \right) = \frac{1}{{2{x^3} - 6{x^2} + 5}} \cr
& {\text{Differentiating}} \cr
& F'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{2{x^3} - 6{x^2} + 5}}} \right] \cr
& {\text{Differentiate by using the formula }}\left( {\frac{f}{h}} \right)' = \frac{{hf' - fh'}}{{{h^2}}},{\text{ }}h \ne 0 \cr
& {\text{Let }}f = 1{\text{ and }}h = 2{x^3} - 6{x^2} + 5,{\text{ then}} \cr
& F'\left( x \right) = \frac{{\left( {2{x^3} - 6{x^2} + 5} \right)\left( 1 \right)' - \left( 1 \right)\left( {2{x^3} - 6{x^2} + 5} \right)'}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}} \cr
& {\text{Compute derivatives}} \cr
& F'\left( x \right) = \frac{{\left( {2{x^3} - 6{x^2} + 5} \right)\left( 0 \right) - \left( {6{x^2} - 12x} \right)}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}} \cr
& {\text{Multiply and simplify}} \cr
& F'\left( x \right) = \frac{{0 - 6{x^2} + 12x}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}} \cr
& F'\left( x \right) = \frac{{12x - 6{x^2}}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}} \cr} $$
