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April 18, 2014

April 18, 2014

Posted by **mary** on Thursday, March 15, 2012 at 1:39am.

I won't say the question because I'd like to figure it out myself or try first, but I'm looking for a simpler way to go about this.

I have to find the derivative of h(x)= (x)(x-1)^1/3

I got: h'(x)= (x-1)^1/3 + (1/3)(x)(x-1)^-2/3

I'm looking for a way to combine the two terms, but the cube roots are messing me up and I'm not completely sure how to go about it... been stuck on this for about an hour...

If any one can help me simplify h'(x) and, if you'd like, show me how, that would be amazing!

Thank you for your time.

- calculus-derivatives -
**Steve**, Thursday, March 15, 2012 at 11:47amNot much joy here. You could always factor out the (x-1)^-2/3 to get

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

= (x-1)^-2/3 * (4x/3 -1)

Depending on what comes next, you may want to massage it some.

- calculus-derivatives -
**mary**, Thursday, March 15, 2012 at 12:04pm:D I'm so happy! Never thought to factor that out x) thank you!

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