Differentiate \(F\left( x \right)\) with respect to \(x\) as follows:
\(\begin{aligned}F'\left( x \right) &= {\left( {4x + 5} \right)^3}\frac{d}{{dx}}\left( {{{\left( {{x^2} - 2x + 5} \right)}^4}} \right) + {\left( {{x^2} - 2x + 5} \right)^4}\frac{d}{{dx}}\left( {{{\left( {4x + 5} \right)}^3}} \right)\\ &= {\left( {4x + 5} \right)^3}\left( {4{{\left( {{x^2} - 2x + 5} \right)}^3}} \right)\frac{d}{{dx}}\left( {{x^2} - 2x + 5} \right) + {\left( {{x^2} - 2x + 5} \right)^4}\\\left( {3{{\left( {4x + 5} \right)}^2}} \right)\frac{d}{{dx}}\left( {4x + 5} \right)\\ &= {\left( {4x + 5} \right)^3}\left( {4{{\left( {{x^2} - 2x + 5} \right)}^3}} \right)\left( {2x - 2} \right) + {\left( {{x^2} - 2x + 5} \right)^4}\left( {3{{\left( {4x + 5} \right)}^2}} \right)\left( 4 \right)\\ &= 4{\left( {4x + 5} \right)^2}{\left( {{x^2} - 2x + 5} \right)^3}\left( {\left( {4x + 5} \right)\left( {2x - 2} \right) + \left( {{x^2} - 2x + 5} \right)\left( 3 \right)} \right)\\ &= 4{\left( {4x + 5} \right)^2}{\left( {{x^2} - 2x + 5} \right)^3}\left( {8{x^2} + 2x - 10 + 3{x^2} - 6x + 15} \right)\\ &= 4{\left( {4x + 5} \right)^2}{\left( {{x^2} - 2x + 5} \right)^3}\left( {11{x^2} - 4x + 5} \right)\end{aligned}\)
Hence, the derivative of \(F\left( x \right)\) is\(4{\left( {4x + 5} \right)^2}{\left( {{x^2} - 2x + 5} \right)^3}\left( {11{x^2} - 4x + 5} \right)\).