Answer
$F'\left( x \right) = 4{\left( {4x + 5} \right)^2}{\left( {{x^2} - 2x + 5} \right)^3}\left( {11{x^2} - 4x + 5} \right)$
Work Step by Step
$$\eqalign{
& F\left( x \right) = {\left( {4x + 5} \right)^3}{\left( {{x^2} - 2x + 5} \right)^4} \cr
& {\text{Differentiate}} \cr
& F'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {4x + 5} \right)}^3}{{\left( {{x^2} - 2x + 5} \right)}^4}} \right] \cr
& {\text{Use the product rule for derivatives}} \cr
& F'\left( x \right) = {\left( {4x + 5} \right)^3}\frac{d}{{dx}}\left[ {{{\left( {{x^2} - 2x + 5} \right)}^4}} \right] \cr
& + {\left( {{x^2} - 2x + 5} \right)^4}\frac{d}{{dx}}\left[ {{{\left( {4x + 5} \right)}^3}} \right] \cr
& {\text{Use the chain rule}} \cr
& F'\left( x \right) = {\left( {4x + 5} \right)^3}\left( 4 \right){\left( {{x^2} - 2x + 5} \right)^3}\frac{d}{{dx}}\left[ {{x^2} - 2x + 5} \right] \cr
& + {\left( {{x^2} - 2x + 5} \right)^4}\left( 3 \right){\left( {4x + 5} \right)^2}\frac{d}{{dx}}\left[ {4x + 5} \right] \cr
& {\text{Computing derivatives}} \cr
& F'\left( x \right) = {\left( {4x + 5} \right)^3}\left( 4 \right){\left( {{x^2} - 2x + 5} \right)^3}\left( {2x - 2} \right) \cr
& + {\left( {{x^2} - 2x + 5} \right)^4}\left( 3 \right){\left( {4x + 5} \right)^2}\left( 4 \right) \cr
& {\text{Multiply}} \cr
& F'\left( x \right) = 4\left( {2x - 2} \right){\left( {4x + 5} \right)^3}{\left( {{x^2} - 2x + 5} \right)^3} \cr
& + 12{\left( {{x^2} - 2x + 5} \right)^4}{\left( {4x + 5} \right)^2} \cr
& {\text{Factor and simplify}} \cr
& F'\left( x \right) = 4{\left( {4x + 5} \right)^2}{\left( {{x^2} - 2x + 5} \right)^3}\left[ {\left( {2x - 2} \right)\left( {4x + 5} \right) + 3\left( {{x^2} - 2x + 5} \right)} \right] \cr
& F'\left( x \right) = 4{\left( {4x + 5} \right)^2}{\left( {{x^2} - 2x + 5} \right)^3}\left[ {8{x^2} + 2x - 10 + 3{x^2} - 6x + 15} \right] \cr
& F'\left( x \right) = 4{\left( {4x + 5} \right)^2}{\left( {{x^2} - 2x + 5} \right)^3}\left( {11{x^2} - 4x + 5} \right) \cr} $$
