Answer
$f'\left( x \right) = 10x{\left( {2{x^3} - 5{x^2} + 4} \right)^4}\left( {3x - 5} \right)$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {2{x^3} - 5{x^2} + 4} \right)^5} \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {2{x^3} - 5{x^2} + 4} \right)}^5}} \right] \cr
& {\text{Use The Power Rule Combined with the Chain Rule}} \cr
& \frac{d}{{dx}}{\left[ {g\left( x \right)} \right]^n} = n{\left[ {g\left( x \right)} \right]^{n - 1}}g'\left( x \right) \cr
& {\text{Therefore}}{\text{,}} \cr
& f'\left( x \right) = 5{\left( {2{x^3} - 5{x^2} + 4} \right)^{5 - 1}}\frac{d}{{dx}}\left[ {2{x^3} - 5{x^2} + 4} \right] \cr
& {\text{Computing derivatives}} \cr
& f'\left( x \right) = 5{\left( {2{x^3} - 5{x^2} + 4} \right)^4}\left( {6{x^2} - 10x} \right) \cr
& {\text{Factoring}} \cr
& f'\left( x \right) = 5\left( {2x} \right){\left( {2{x^3} - 5{x^2} + 4} \right)^4}\left( {3x - 5} \right) \cr
& f'\left( x \right) = 10x{\left( {2{x^3} - 5{x^2} + 4} \right)^4}\left( {3x - 5} \right) \cr} $$
