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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    I know its quotient rule but i can't seem to get it to work out

  2. anonymous
    • 5 years ago
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    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. :)

  3. anonymous
    • 5 years ago
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    For the second derivative, yeah, you shouldn't need quotient rule...you can continue using chain rule to your heart's delight.

  4. anonymous
    • 5 years ago
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    Ohhhhh, didn't see the x before the fraction...sorry. :P

  5. anonymous
    • 5 years ago
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    \[\sqrt[3]{x}/(1-x)\]

  6. anonymous
    • 5 years ago
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    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.

  7. anonymous
    • 5 years ago
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    ...it's going to become very ugly, very quickly. :P

  8. anonymous
    • 5 years ago
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    I know! I've tried working it out multiple times times with no succes

  9. anonymous
    • 5 years ago
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    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.

  10. anonymous
    • 5 years ago
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    ...let me know if you see any careless mistakes

  11. anonymous
    • 5 years ago
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    I get the same thing

  12. anonymous
    • 5 years ago
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    Okay, now...get your self a biiiig piece of paper...and churn out the second. :D

  13. anonymous
    • 5 years ago
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    can you also write the first as [1-x+x^(1/3)]/ [3x^(2/3)(1-x)^2?

  14. anonymous
    • 5 years ago
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    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!

  15. anonymous
    • 5 years ago
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    thank you all!

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