Answer
$$\frac{{dy}}{{dx}} = 18x{\left( {2x + 1} \right)^2} + 3{\left( {2x + 1} \right)^3}$$
Work Step by Step
$$\eqalign{
& y = 3x{\left( {2x + 1} \right)^3} \cr
& {\text{differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {3x{{\left( {2x + 1} \right)}^3}} \right] \cr
& {\text{use product rule}} \cr
& \frac{{dy}}{{dx}} = 3x\frac{d}{{dx}}\left[ {{{\left( {2x + 1} \right)}^3}} \right] + {\left( {2x + 1} \right)^3}\frac{d}{{dx}}\left[ {3x} \right] \cr
& {\text{use the power rule with the chain rule }}\frac{d}{{dx}}\left[ {g{{\left( x \right)}^n}} \right] = ng{\left( x \right)^{n - 1}}\frac{d}{{dx}}\left[ {g'\left( x \right)} \right]{\text{ then}} \cr
& \frac{{dy}}{{dx}} = 3x\left( 3 \right){\left( {2x + 1} \right)^3}\frac{d}{{dx}}\left[ {2x + 1} \right] + {\left( {2x + 1} \right)^3}\frac{d}{{dx}}\left[ {3x} \right] \cr
& {\text{find derivatives}} \cr
& \frac{{dy}}{{dx}} = 3x\left( 3 \right){\left( {2x + 1} \right)^2}\left( 2 \right) + {\left( {2x + 1} \right)^3}\left( 3 \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = 18x{\left( {2x + 1} \right)^2} + 3{\left( {2x + 1} \right)^3} \cr} $$
