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iplayffxiv

  • one year ago

Find the second derivative of the function. s=(t^6+3t+3)/ t^2

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  1. Mertsj
    • one year ago
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    It might be easier to write it like this: \[s=t^4+3t ^{-1}+3t ^{-2}\]

  2. sasogeek
    • one year ago
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    @iplayffxiv can you continue from the help @Mertsj has provided?

  3. iplayffxiv
    • one year ago
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    i cant

  4. iplayffxiv
    • one year ago
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    i dont understand that

  5. sasogeek
    • one year ago
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    He simplified the equation first. now it will be easier to find the derivatives :) still lost? here's a simple way to find a derivative: if \(\large y=ax^n \) the derivative \(\large dy \), is equal to \(\large dy=nax^{n-1} \)

  6. sasogeek
    • one year ago
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    \(\large a \) is a constant, if \(\large a \) is 1, it is not written

  7. iplayffxiv
    • one year ago
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    does not make sense to me. it is multiple choice. can you choose the right answer?

  8. sasogeek
    • one year ago
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    first derivative: \(\large s=4t^3-3t^{-2}-6t^{-3} \) apply the same trick to see if you get ur answer.... take ur time, shouldn't be that difficult :)

  9. iplayffxiv
    • one year ago
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    A) d2s/dt2 = 4t^3 -3/t2 -6/t3 B) d2s/dt2 = 12t^4 +6/t +18/t2 C) d2s/dt2 = 4t^2 -3/t3 -6/t4 D) d2s/dt2 = 12t^2 +6/t^3 +18/t4

  10. iplayffxiv
    • one year ago
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    you see?

  11. sasogeek
    • one year ago
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    find the derivative of \(\large 4t^3 \), your answer will be obvious :) remember, the derivative of \(\large ax^n =nax^{n-1} \)

  12. iplayffxiv
    • one year ago
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    i dont know what to do

  13. sasogeek
    • one year ago
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    do well to read on how to find derivatives, but the answer you're looking for is the last option

  14. iplayffxiv
    • one year ago
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    someone give me answer

  15. Mertsj
    • one year ago
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    Do you want to learn to do this or do you just want an answer?

  16. iplayffxiv
    • one year ago
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    i would like to have the answer

  17. Mertsj
    • one year ago
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    Then we are all leaving because OS is not an answering service. Goodbye and good luck. All the time that has been spent on this problem was ample for you to learn to do i had you so chosen.

  18. Loser66
    • one year ago
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    good luck bye!!!

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