Derivatives
posted by Erika on .
Find f'(x): (abs(((x^2)*((3x+2) ^(1/3)))/((2x3)^3))
So I thought I should split it into two parts: A and B.
A) (x^2)/((2x3)^2)
B) ((3x+2)^(1/3))/(2x3)
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)/((2x3)^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:
((3x+2)^(2/3)*(2x3)  4(x3)*((3x+2)^(1/3)))/((2x3)^2)
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!

X^2(3x2)^2/3/(2x3)^3  2x(x+3)(3x+2)^1/3/(2x3)^4

The strategy is interesting, since you'll have to apply the quotient rule to each part, then the product rule.
Your value in A is correct, but note that
(12x^2 + 18x)/((2x3)^4)
= 6x(2x3)/(2x3)^4
= 6x/(2x3)^3
B' = (4x+7)/((2x3)^2 (3x+2)^(2/3))
Now you have to evaluate
A'B + AB'
What a mess.
I'd rather just work with the original expression. Unfortunately, the parentheses are not balanced. Try
f(x) = uv/w^3 if that is what is meant. Let us know what u,v,w are. I think it's clear, but I'm not sure where the  applies.