Answer
$G'\left( u \right) = \frac{{18{u^4} + 24{u^3} - 5}}{{{{\left( {u + 1} \right)}^2}}}$
Work Step by Step
$$\eqalign{
& G\left( u \right) = \frac{{6{u^4} - 5u}}{{u + 1}} \cr
& {\text{Differentiating}} \cr
& G'\left( u \right) = \frac{d}{{du}}\left[ {\frac{{6{u^4} - 5u}}{{u + 1}}} \right] \cr
& {\text{Differentiate by using the formula }}\left( {\frac{f}{h}} \right)' = \frac{{hf' - fh'}}{{{h^2}}},{\text{ }}h \ne 0 \cr
& {\text{Let }}f = 6{u^4} - 5u{\text{ and }}h = u + 1,{\text{ then}} \cr
& G'\left( u \right) = \frac{{\left( {u + 1} \right)\left( {6{u^4} - 5u} \right)' - \left( {6{u^4} - 5u} \right)\left( {u + 1} \right)'}}{{{{\left( {u + 1} \right)}^2}}} \cr
& {\text{Compute derivatives}} \cr
& G'\left( u \right) = \frac{{\left( {u + 1} \right)\left( {24{u^3} - 5} \right) - \left( {6{u^4} - 5u} \right)\left( 1 \right)}}{{{{\left( {u + 1} \right)}^2}}} \cr
& {\text{Multiply and simplify}} \cr
& G'\left( u \right) = \frac{{24{u^4} - 5u + 24{u^3} - 5 - 6{u^4} + 5u}}{{{{\left( {u + 1} \right)}^2}}} \cr
& G'\left( u \right) = \frac{{18{u^4} + 24{u^3} - 5}}{{{{\left( {u + 1} \right)}^2}}} \cr} $$
