Answer
$$n'\left( x \right) = \frac{{ - 87x - 95}}{{{{\left( {29x + 7} \right)}^3}}}$$
Work Step by Step
$$\eqalign{
& n\left( x \right) = \left( {3x + 2} \right){\left( {29x + 7} \right)^{ - 2}} \cr
& {\text{use }}{a^{ - n}} = \frac{1}{{{a^n}}} \cr
& n\left( x \right) = \frac{{3x + 2}}{{{{\left( {29x + 7} \right)}^2}}} \cr
& {\text{Calculate the derivative of the function}} \cr
& n'\left( x \right) = \frac{d}{{dx}}\left( {\frac{{3x + 2}}{{{{\left( {29x + 7} \right)}^2}}}} \right) \cr
& {\text{use quotient rule}} \cr
& n'\left( x \right) = \frac{{{{\left( {29x + 7} \right)}^2}\left( {3x + 2} \right)' - \left( {3x + 2} \right)\left( {{{\left( {29x + 7} \right)}^2}} \right)'}}{{{{\left( {29x + 7} \right)}^4}}} \cr
& n'\left( x \right) = \frac{{{{\left( {29x + 7} \right)}^2}\left( 3 \right) - \left( {3x + 2} \right)\left( {2\left( {29x + 7} \right)\left( {29} \right)} \right)}}{{{{\left( {29x + 7} \right)}^4}}} \cr
& {\text{simplifying}} \cr
& n'\left( x \right) = \frac{{\left( {29x + 7} \right)\left( 3 \right) - \left( {3x + 2} \right)\left( {2\left( {29} \right)} \right)}}{{{{\left( {29x + 7} \right)}^3}}} \cr
& n'\left( x \right) = \frac{{87x + 21 - 174x - 116}}{{{{\left( {29x + 7} \right)}^3}}} \cr
& n'\left( x \right) = \frac{{ - 87x - 95}}{{{{\left( {29x + 7} \right)}^3}}} \cr} $$
