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iplayffxiv

  • 2 years ago

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

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

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

  3. iplayffxiv
    • 2 years ago
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    i cant

  4. iplayffxiv
    • 2 years ago
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    i dont understand that

  5. sasogeek
    • 2 years 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
    • 2 years ago
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    \(\large a \) is a constant, if \(\large a \) is 1, it is not written

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

  8. sasogeek
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    you see?

  11. sasogeek
    • 2 years 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
    • 2 years ago
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    i dont know what to do

  13. sasogeek
    • 2 years 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
    • 2 years ago
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    someone give me answer

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

  16. iplayffxiv
    • 2 years ago
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    i would like to have the answer

  17. Mertsj
    • 2 years 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
    • 2 years ago
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    good luck bye!!!

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