find the first and second derivative of cuberootx/ (1-x)

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find the first and second derivative of cuberootx/ (1-x)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I know its quotient rule but i can't seem to get it to work out
Okay, so you can re-write it as (1-x) ^ 1/3. Use chain rule once to get 1/3 * -1 * (1-x) ^ (1/3 - 1). Lather, rinse, repeat. :)
For the second derivative, yeah, you shouldn't need quotient rule...you can continue using chain rule to your heart's delight.

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Ohhhhh, didn't see the x before the fraction...sorry. :P
\[\sqrt[3]{x}/(1-x)\]
I see now. You can assign a general term to the numerator, and to the denominator -- like u, and v, respectively. The quotient rule goes something like d/dx (u/v) = ( v * u' - u * v') / v^2.
...it's going to become very ugly, very quickly. :P
I know! I've tried working it out multiple times times with no succes
Alright, I understand where it's easy to go wrong here...it's a lot of terms being tossed about, and to keep track of; let's try going through the first derivative, and you can try to do the second (give yourself some space!). d/dx (cbrt(x) / (1-x) ) = ( (1-x) * 1/3 * x^(-2/3) - cbrt(x) * -1 ) / (1-x)^2.
...let me know if you see any careless mistakes
I get the same thing
Okay, now...get your self a biiiig piece of paper...and churn out the second. :D
can you also write the first as [1-x+x^(1/3)]/ [3x^(2/3)(1-x)^2?
looks good! and the best part is, i have to find the zeros of both because i have use that information to draw a graph. whoohoo!
thank you all!

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