i do know that one of the rules involving natural logarithms is: y=lnx which derived is y'= 1/x . But I'mstill not sure. This is my teacher's solution:y'= -5-3x^2 / (-5-5x-x^3)
y =ln(-5-5x-x^3)
derivative of ln(x) is y' = 1/(x)*(x)'
y'= (ln(-5-5x-x^3))'*(-5-5x-x^3)' so, you have 1/(-5-5x-x^3)*(5x-3x^2) =(5x-3x^2)/(-5-5x-x^3)
dokrbb