Derivative of Cube Root (1/(2-3x))

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SUMMARY

The derivative of the cube root of the function \( \frac{1}{2-3x} \) is calculated using the chain rule. The correct derivative is \( \frac{1}{(2-3x)^{4/3}} \). The initial attempt at the solution involved transforming the cube root to an exponent of \( \frac{1}{3} \) and resulted in a more complex expression that included \( \frac{1}{((1/(2-3x))^{2/3}(2-3x)^2)} \). The confusion arose from the simplification process and the use of Maple software for verification.

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Homework Statement


Find the derivative of: cube root of (1/(2-3x))



Homework Equations


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



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!
 
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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?
 
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!
 

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