anonymous
  • anonymous
find the first and second derivative of cuberootx/ (1-x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I know its quotient rule but i can't seem to get it to work out
anonymous
  • anonymous
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. :)
anonymous
  • anonymous
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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anonymous
  • anonymous
Ohhhhh, didn't see the x before the fraction...sorry. :P
anonymous
  • anonymous
\[\sqrt[3]{x}/(1-x)\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
...it's going to become very ugly, very quickly. :P
anonymous
  • anonymous
I know! I've tried working it out multiple times times with no succes
anonymous
  • anonymous
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.
anonymous
  • anonymous
...let me know if you see any careless mistakes
anonymous
  • anonymous
I get the same thing
anonymous
  • anonymous
Okay, now...get your self a biiiig piece of paper...and churn out the second. :D
anonymous
  • anonymous
can you also write the first as [1-x+x^(1/3)]/ [3x^(2/3)(1-x)^2?
anonymous
  • anonymous
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!
anonymous
  • anonymous
thank you all!

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