Answer
$$m'\left( t \right) = - \frac{{{e^{ - 5t}}\left( {30t + 8} \right)}}{{{{\left( {6t - 2} \right)}^4}}}$$
Work Step by Step
$$\eqalign{
& m\left( t \right) = \frac{{{e^{ - 5t}}}}{{{{\left( {6t - 2} \right)}^3}}} \cr
& {\text{Calculate the derivative of the function}} \cr
& m'\left( t \right) = \frac{d}{{dt}}\left( {\frac{{{e^{ - 5t}}}}{{{{\left( {6t - 2} \right)}^3}}}} \right) \cr
& {\text{use quotient rule}} \cr
& m'\left( t \right) = \frac{{{{\left( {6t - 2} \right)}^3}\left( {{e^{ - 5t}}} \right)' - {e^{ - 5t}}\left( {{{\left( {6t - 2} \right)}^3}} \right)'}}{{{{\left( {{{\left( {6t - 2} \right)}^3}} \right)}^2}}} \cr
& {\text{Compute derivatives}} \cr
& m'\left( t \right) = \frac{{{{\left( {6t - 2} \right)}^3}\left( {{e^{ - 5t}}} \right)\left( { - 5} \right) - {e^{ - 5t}}\left( {3{{\left( {6t - 2} \right)}^2}\left( 6 \right)} \right)}}{{{{\left( {6t - 2} \right)}^6}}} \cr
& {\text{simplifying}} \cr
& m'\left( t \right) = \frac{{\left( {6t - 2} \right)\left( {{e^{ - 5t}}} \right)\left( { - 5} \right) - {e^{ - 5t}}\left( {3\left( 6 \right)} \right)}}{{{{\left( {6t - 2} \right)}^4}}} \cr
& m'\left( t \right) = \frac{{{e^{ - 5t}}\left( { - 30t + 10 - 18} \right)}}{{{{\left( {6t - 2} \right)}^4}}} \cr
& m'\left( t \right) = - \frac{{{e^{ - 5t}}\left( {30t + 8} \right)}}{{{{\left( {6t - 2} \right)}^4}}} \cr} $$
