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anonymous

  • one year ago

I need help in calculus. Can anyone?

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  1. amistre64
    • one year ago
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    calculus 1? hmm

  2. amistre64
    • one year ago
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    posting the question might be prudent

  3. anonymous
    • one year ago
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    Sorry for a late reply. Can you help me find the derivative of this y=-3/5cot^5 x/3+cot^3 x/3-3cot x/3 -x?

  4. amistre64
    • one year ago
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    using derivative rules or the other way?

  5. amistre64
    • one year ago
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    is that cot^5 a top or bottom of the fraction?

  6. anonymous
    • one year ago
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    Using the rules. It's on top.

  7. amistre64
    • one year ago
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    are those x/3 arguements of the trig function, or is that trig divided by 3?

  8. amistre64
    • one year ago
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    \[y=-\frac35\cot^5 (\frac{x}3)+\cot^3 (\frac{x}3)-3\cot (\frac{x}3) -x\]

  9. anonymous
    • one year ago
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    Yes that's it!

  10. amistre64
    • one year ago
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    if we let u=cot(x/3) \[y=-\frac35u^5 +u^3 -3u -x\] how would you attempt to find the derivative of this?

  11. amistre64
    • one year ago
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    what is your power rule?

  12. anonymous
    • one year ago
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    nu^n-1 du/dx

  13. amistre64
    • one year ago
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    good, lets use u' instead of du/dx just simpler to write

  14. amistre64
    • one year ago
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    \[y=-\frac35u^5 +u^3 -3u -x\] \[y'=-\frac3u^4u' +3u^2u' -u' -x'\] x'=dx/dx=1 soo \[y'=u'(-\frac3u^4 +3u^2 -1) -1\] oh, and what will our u' be since u=cot(x/3)?

  15. amistre64
    • one year ago
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    bah, code typo .... -3u^4

  16. amistre64
    • one year ago
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    u' is a chain rule as well ... what is your chain rule?

  17. anonymous
    • one year ago
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    nu^n-1 du/dx

  18. amistre64
    • one year ago
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    no, thats power rule .... chain rule is for composition of function

  19. amistre64
    • one year ago
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    a function within a function, within a function, within a function .... f(g(h(...(k(x))))) the derivative is a multiplier effect: f'g'h'...k'

  20. anonymous
    • one year ago
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    3/5cot^5(x/3) will be like this 3/5 (5)cot^4(-x/9)(-csc^2x)?

  21. amistre64
    • one year ago
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    close, but i like to work with a simpler writeup 3/5 u^5 would be 3u^4 u' but since u = cot(x/3), u' = -csc^2(x/3) * (x/3)'; or -1/3 csc^2(x/3)

  22. anonymous
    • one year ago
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    Really going to be a tough one for me.

  23. amistre64
    • one year ago
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    let u = cot(x/3); u' = -1/3 csc^2(x/3) \[y'=u'(-3u^4 +3u^2 -1) -1\] \[y'=-\frac13\csc^2(\frac{x}3)~(-3\cot^4(\frac{x}3) +3\cot^2(\frac{x}3) -1) -1\] or \[y'=\csc^2(\frac{x}3)~(\cot^4(\frac{x}3) -\cot^2(\frac{x}3) +\frac13) ~-1\]

  24. amistre64
    • one year ago
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    its not a difficult process, just messy if you try to keep it all ... x-ish thru the process

  25. anonymous
    • one year ago
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    I'll try for it. This is, one part of Differential Calculus that I'm really having a hard time.

  26. amistre64
    • one year ago
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    it can be daunting, but its really simpler than algebra :) good luck

  27. anonymous
    • one year ago
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    Maybe but I don't really know. Thanks.

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