Answer
$$f'\left( x \right) = 2\left( 2 \right){\left( {2x + 3} \right)^{2 - 1}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {2x + 3} \right)^2} \cr
& {\text{Expanding the binomial}} \cr
& f\left( x \right) = {\left( {2x} \right)^2} + 2\left( {2x} \right)\left( 3 \right) + {\left( 3 \right)^2} \cr
& f\left( x \right) = 4{x^2} + 12x + 9 \cr
& {\text{Differentiate by using the rule }}\frac{d}{{dx}}\left[ {{x^r}} \right] = r{x^{r - 1}} \cr
& f'\left( x \right) = 8x + 12 \cr
& {\text{Factoring}} \cr
& f'\left( x \right) = 4\left( {2x + 3} \right) \cr
& or \cr
& f'\left( x \right) = 2\left( 2 \right){\left( {2x + 3} \right)^{2 - 1}} \cr
& {\text{We have verified that for }} \cr
& f\left( x \right) = {\left( {2x + 3} \right)^2} \Rightarrow f'\left( x \right) = 2\left( 2 \right){\left( {2x + 3} \right)^{2 - 1}} \cr
& f\left( x \right) = {\left( {mx + b} \right)^n}\,\, \Rightarrow \,\,f\left( x \right) = nm{\left( {mx + b} \right)^{n - 1}}\,\, \cr} $$
