Answer
$g'\left( t \right) = \frac{{ - 17}}{{{{\left( {5t + 1} \right)}^2}}}$
Work Step by Step
$$\eqalign{
& g\left( t \right) = \frac{{3 - 2t}}{{5t + 1}} \cr
& {\text{Differentiating}} \cr
& g'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{3 - 2t}}{{5t + 1}}} \right] \cr
& {\text{Differentiate by using the formula }}\left( {\frac{f}{h}} \right)' = \frac{{hf' - fh'}}{{{h^2}}},{\text{ }}h \ne 0 \cr
& {\text{Let }}f = 3 - 2t{\text{ and }}h = 5t + 1,{\text{ then}} \cr
& g'\left( t \right) = \frac{{\left( {5t + 1} \right)\left( {3 - 2t} \right)' - \left( {3 - 2t} \right)\left( {5t + 1} \right)'}}{{{{\left( {5t + 1} \right)}^2}}} \cr
& {\text{Compute derivatives}} \cr
& g'\left( t \right) = \frac{{\left( {5t + 1} \right)\left( { - 2} \right) - \left( {3 - 2t} \right)\left( 5 \right)}}{{{{\left( {5t + 1} \right)}^2}}} \cr
& {\text{Multiply and simplify}} \cr
& g'\left( t \right) = \frac{{ - 10t - 2 - 15 + 10t}}{{{{\left( {5t + 1} \right)}^2}}} \cr
& g'\left( t \right) = \frac{{ - 17}}{{{{\left( {5t + 1} \right)}^2}}} \cr} $$
