Algebraic problem with roots and indices: derivative of h(x) = cbrt{x^3 - x^2}

[Phenomniverse]

Can anyone tell me whether further playing around with this one is going to yield a more simplified result? I'm at a bit of a dead end with it. See attached pic.

 

[mmm4444bot]

Unless you were instructed to rationalize denominators, you could stop at

\(\displaystyle \dfrac{3x^2 - 2x}{3 \cdot \sqrt[3]{(x^3-x^2)^2}}\)


Some instructors would accept

\(\displaystyle h^{\prime}(x) = \dfrac{1}{3} \cdot (x^3 - x^2)^{-2/3} \cdot (3x^2 - 2x)\)

 

[Phenomniverse]

Great, thanks

 

[Phenomniverse]

What about this:

Is this a legitimate series of operations?

 

[Phenomniverse]

Correction, the last three lines should have +x^4, +x and +x as their last terms under the cube root.

 

[mmm4444bot]

… Is this a legitimate series of operations?

Divide line 4 by the corrected line 8. If you get 1, then you've confirmed they are equivalent. :cool:

 

[Phenomniverse]

Ok, well first of all, here is the corrected version of the above simplification, with one additional step:


And here is my proof that the fourth line divided by the final line is equal to 1:

The problem with this method of double checking though, is that it involves doing the same operations as the original simplification, so it doesn't really constitute proof that those operations are legitimate.