anonymous
  • anonymous
Find the second derivative of the function. s=(t^6+3t+3)/ t^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mertsj
  • Mertsj
It might be easier to write it like this: \[s=t^4+3t ^{-1}+3t ^{-2}\]
sasogeek
  • sasogeek
@iplayffxiv can you continue from the help @Mertsj has provided?
anonymous
  • anonymous
i cant

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anonymous
  • anonymous
i dont understand that
sasogeek
  • sasogeek
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 \(\large dy=nax^{n-1} \)
sasogeek
  • sasogeek
\(\large a \) is a constant, if \(\large a \) is 1, it is not written
anonymous
  • anonymous
does not make sense to me. it is multiple choice. can you choose the right answer?
sasogeek
  • sasogeek
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 :)
anonymous
  • anonymous
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
anonymous
  • anonymous
you see?
sasogeek
  • sasogeek
find the derivative of \(\large 4t^3 \), your answer will be obvious :) remember, the derivative of \(\large ax^n =nax^{n-1} \)
anonymous
  • anonymous
i dont know what to do
sasogeek
  • sasogeek
do well to read on how to find derivatives, but the answer you're looking for is the last option
anonymous
  • anonymous
someone give me answer
Mertsj
  • Mertsj
Do you want to learn to do this or do you just want an answer?
anonymous
  • anonymous
i would like to have the answer
Mertsj
  • Mertsj
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.
Loser66
  • Loser66
good luck bye!!!

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