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## anonymous 5 years ago find the first and second derivative of cuberootx/ (1-x)

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1. anonymous

I know its quotient rule but i can't seem to get it to work out

2. 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. :)

3. anonymous

For the second derivative, yeah, you shouldn't need quotient rule...you can continue using chain rule to your heart's delight.

4. anonymous

Ohhhhh, didn't see the x before the fraction...sorry. :P

5. anonymous

$\sqrt[3]{x}/(1-x)$

6. 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.

7. anonymous

...it's going to become very ugly, very quickly. :P

8. anonymous

I know! I've tried working it out multiple times times with no succes

9. 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.

10. anonymous

...let me know if you see any careless mistakes

11. anonymous

I get the same thing

12. anonymous

Okay, now...get your self a biiiig piece of paper...and churn out the second. :D

13. anonymous

can you also write the first as [1-x+x^(1/3)]/ [3x^(2/3)(1-x)^2?

14. 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!

15. anonymous

thank you all!

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