Answer
$y' = \frac{{2x + 3}}{{\left( {{x^2} + 3x} \right)\left( {\ln 8} \right)}}$
Work Step by Step
$$\eqalign{
& y = {\log _8}\left( {{x^2} + 3x} \right) \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {{{\log }_8}\left( {{x^2} + 3x} \right)} \right] \cr
& {\text{Use the formula }}\frac{d}{{dx}}\left( {{{\log }_b}x} \right) = \frac{1}{{x\ln b}}{\text{ and the chain rule}} \cr
& y' = \frac{1}{{\left( {{x^2} + 3x} \right)\left( {\ln 8} \right)}}\frac{d}{{dx}}\left[ {{x^2} + 3x} \right] \cr
& {\text{compute the derivative and simplify}} \cr
& y' = \frac{1}{{\left( {{x^2} + 3x} \right)\left( {\ln 8} \right)}}\left( {2x + 3} \right) \cr
& y' = \frac{{2x + 3}}{{\left( {{x^2} + 3x} \right)\left( {\ln 8} \right)}} \cr} $$
