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