Answer to Question #204092 in Calculus for moe

Question

What is the derivative of ln$( (x^3 - 1) / x^2 )$

Answer 1

$$y=\ln(\dfrac{x^3-1}{x^2})$$

$$x^3-1>0=>x>1$$

Then

$$y=\ln(\dfrac{x^3-1}{x^2})=\ln(x^3-1)-2\ln x$$

$$y'=\ln(\dfrac{x^3-1}{x^2})'=(\ln(x^3-1)-2\ln x)'$$

$$=\dfrac{3x^2}{x^3-1}-\dfrac{2}{x}=\dfrac{x^3+2}{x(x^3-1)}$$