Find f'(x): (abs(((x^2)*((3x+2) ^(1/3)))/((2x-3)^3))

Question

So I thought I should split it into two parts: A and B.

A) (x^2)/((2x-3)^2) 
B) ((3x+2)^(1/3))/(2x-3)

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:

((3x+2)^(-2/3)*(2x-3) - 4(x-3)*((3x+2)^(1/3)))/((2x-3)^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!

Kuai · Answer

X^2(3x-2)^-2/3/(2x-3)^3 - 2x(x+3)(3x+2)^1/3/(2x-3)^4

Steve · Answer

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)/((2x-3)^4)
= -6x(2x-3)/(2x-3)^4
= -6x/(2x-3)^3

B' = -(4x+7)/((2x-3)^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.