Answer
$$\frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{{{\left( {4x + 3} \right)}^3}\left( {12x - 3} \right)}}{{{x^2}}}$$
Work Step by Step
$$\eqalign{
& C\left( x \right) = {\left( {4x + 3} \right)^4} \cr
& {\text{The marginal average cost is the derivative of the average cost function}}{\text{. then}} \cr
& {\text{the marginal average cost is given by}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{d}{{dx}}\left[ {\frac{{{{\left( {4x + 3} \right)}^4}}}{x}} \right] \cr
& {\text{by using the quotient rule}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{x\frac{d}{{dx}}\left[ {{{\left( {4x + 3} \right)}^4}} \right] - {{\left( {4x + 3} \right)}^4}\frac{d}{{dx}}\left[ x \right]}}{{{x^2}}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{x\left( 4 \right){{\left( {4x + 3} \right)}^3}\left( 4 \right) - {{\left( {4x + 3} \right)}^4}\left( 1 \right)}}{{{x^2}}} \cr
& {\text{simplify}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{16x{{\left( {4x + 3} \right)}^3} - {{\left( {4x + 3} \right)}^4}}}{{{x^2}}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{{{\left( {4x + 3} \right)}^3}\left( {16x - 4x - 3} \right)}}{{{x^2}}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{{{\left( {4x + 3} \right)}^3}\left( {12x - 3} \right)}}{{{x^2}}} \cr} $$
