# Derivative question?

1. Oct 6, 2008

### ashleyrc

1. The problem statement, all variables and given/known data
Find the derivative of: cube root of (1/(2-3x))

2. Relevant equations
use chain rule (i think?):(f of g)'(x) = f'(g(x))*g'(x)

3. The attempt at a solution
first, i changed the cube root to the exponent 1/3.
second, i did the derivative and came up with 1/((1/(2-3*x))^(2/3)*(2-3*x)^2)
now i think i need to cancel out the 1/(2-3*x))^(2/3), but i'm not sure how to go about it.
---i kinda cheated and went on maple to see if my answer was close, and it came up as 1/(9*x^2-12*x+4), and i knew i had to factor the last part of the denominator, but i don't understand how the first part of the denominator disappeared.

thanks for all of your help!

2. Oct 6, 2008

### Dick

Your answer is more correct than maple's answer. Are you sure you typed it in correctly? You should get 1/(2-3x)^(4/3). 2-2/3=4/3. Sure you can't see how the exponents add up to that?

3. Oct 7, 2008

### ashleyrc

yeah, i guess that makes sense. the division signs in the equation tend to confuse me. i'll have to do some more problems before i get the hang of it. it's still pretty confusing.
i must've used maple wrong. i've only started using it the past week. it seems like i need a degree just to understand how to use it :)
Thank you for the help!!!