1. The problem statement, all variables and given/known data Find f'(x): [itex] (abs(((x^2)*((3x+2) ^(1/3)))/((2x-3)^3)) [/itex] <- Not sure why it''s not showing up but the 1/3 is an exponent to just the (3x+2). 2. Relevant equations Product Rule and Quotient Rule for Differentiating 3. The attempt at a solution So I thought I should split it into two parts: A and B. A) [itex] (x^2)/((2x-3)^2) [/itex] B) [itex] ((3x+2)^1/3)/(2x-3) [/itex] My plan was to differentiate each part and then multiply them together, since multiplying their denominators together gives me the original denominator power (4). I end up getting for A) (-12x^2 + 18x)/((2x-3)^4) For B, I have something which seems much more complex, because of the negative fractional exponents, and this is where I'm stuck. I want to simplify, but I'm not sure how to without screwing everything up. This is my last step for B: [itex] ((3x+2)^(-2/3)*(2x-3) - 4(x-3)*((3x+2)^(1/3)))/((2x-3)^2) [/itex] <- The -2/3 is the exponent I really appreciate the help. And also, as for the absolute value part of the question, well I haven't even thought about that- I'm stuck on the easy part itself. -_- Thank you!