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anonymous

  • one year ago

As soon as I think I've finished, more work appears! YAY! (EXTREME sarcasm). I'll give a medal to anyone who can help me with the problem I'm posting below. :)

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  1. anonymous
    • one year ago
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    \[(x^2-1)^\frac{5}{2} (x^3+5)\]

  2. jim_thompson5910
    • one year ago
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    what are the instructions?

  3. anonymous
    • one year ago
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    Sorry, the directions are to find the derivative. (not sure why i didn't post that?)

  4. anonymous
    • one year ago
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    So far I've gotten...

  5. anonymous
    • one year ago
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    \[\:\left(x^2-1\right)^{\frac{5}{2}}\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\left(x^2-1\right)^{\frac{3}{2}}\left(2x\right)\]

  6. jim_thompson5910
    • one year ago
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    it looks great

  7. anonymous
    • one year ago
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    I'm not exactly sure what I should do next. Should I begin to simplify or is there another step involved?

  8. jim_thompson5910
    • one year ago
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    there's not much to simplify after this point

  9. anonymous
    • one year ago
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    So that's it?

  10. jim_thompson5910
    • one year ago
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    the only thing I can see really is factoring but it's not much of a simplification if you ask me

  11. anonymous
    • one year ago
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    My teacher prefers specifics unfortunately. Could you help me identify what parts should be factored?

  12. jim_thompson5910
    • one year ago
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    notice there's a x^2-1 repeated

  13. jim_thompson5910
    • one year ago
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    \[\large \left(x^2-1\right)^{\frac{5}{2}}\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\left(x^2-1\right)^{\frac{3}{2}}\left(2x\right)\] \[\large \color{red}{\left(x^2-1\right)}^{\frac{5}{2}}\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\color{red}{\left(x^2-1\right)}^{\frac{3}{2}}\left(2x\right)\]

  14. jim_thompson5910
    • one year ago
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    Factor it out to get \[\large \color{red}{\left(x^2-1\right)}^{\frac{5}{2}}\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\color{red}{\left(x^2-1\right)}^{\frac{3}{2}}\left(2x\right)\] \[\large \color{red}{\left(x^2-1\right)}^{\frac{3}{2}}\left(\left(x^2-1\right)\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\left(2x\right)\right)\] as you can see, it's not much of a simplification

  15. jim_thompson5910
    • one year ago
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    well now that I think about it more, the 5/2 and 2x multiply to 5x so that's something

  16. anonymous
    • one year ago
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    Question: How come the 3/2 exponent was kept but not the 5/2 ?

  17. jim_thompson5910
    • one year ago
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    I factored out the portion with the 3/2 exponent

  18. jim_thompson5910
    • one year ago
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    |dw:1444358829913:dw|

  19. jim_thompson5910
    • one year ago
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    |dw:1444358856055:dw|

  20. jim_thompson5910
    • one year ago
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    |dw:1444358869946:dw|

  21. jim_thompson5910
    • one year ago
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    that's a similar example

  22. anonymous
    • one year ago
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    Okay, I think I understand. Is that the only factoring that can be done?

  23. jim_thompson5910
    • one year ago
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    yeah I think so, but again, it's not much of a simplification

  24. jim_thompson5910
    • one year ago
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    I guess you could do this \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(\left(x^2-1\right)\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\left(2x\right)\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left((x^2-1)(3x^2)+(x^3+5)(5x)\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(3x^3-3x^2+5x^4+25x\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(35x^4+x^3-3x^2+25x\right)\]

  25. jim_thompson5910
    • one year ago
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    sorry typo, let me fix

  26. jim_thompson5910
    • one year ago
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    \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(\left(x^2-1\right)\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\left(2x\right)\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left((x^2-1)(3x^2)+(x^3+5)(5x)\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(3x^4-3x^2+5x^4+25x\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(8x^4-3x^2+25x\right)\]

  27. anonymous
    • one year ago
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    Oh, that's perfect! Thank you yet again!

  28. anonymous
    • one year ago
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    Now, I am off to learn about Extrema Derivatives. Wish me luck!

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