Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

LolWolf

Does anyone have any nice ways of handling differentials like \[ \frac{d^2}{dx^2}\left(\frac{20}{1+x^2}\right)\]? And, the more extreme cases like \[\frac{d^4}{dx^4}\left(\frac{20}{1+x^2}\right)\].

  • one year ago
  • one year ago

  • This Question is Closed
  1. eSpeX
    Best Response
    You've already chosen the best response.
    Medals 0

    Have you considered using a negative exponent? \[\frac{d^2}{dx^2}(20)(1+x^2)^{-1}\]

    • one year ago
  2. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, but the problem gets ugly, quickly, especially considering we are taking the second derivative. So, I'm guessing, there is none? All just straight up computation. Hmm.

    • one year ago
  3. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 0

    i usually use a "u-substitution" in derivatives...i suppose that makes me weird....

    • one year ago
  4. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    Enlighten me...?

    • one year ago
  5. calculusfunctions
    Best Response
    You've already chosen the best response.
    Medals 0

    @LolWolf \[\frac{ d ^{2} }{ dx ^{2} }(\frac{ 20 }{ 1+x ^{2} })\]implies that you find the second derivative of the function y, if\[y =\frac{ 20 }{ 1+x ^{2} }\]I assume you know your derivative rules, correct?

    • one year ago
  6. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, I'm not wondering so much how to find it, but, rather, how to do it neatly.

    • one year ago
  7. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    (If someone has some ingenious way of managing it, et al)

    • one year ago
  8. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    y'=-40x/(1+x^2)^2 now do it again the second time

    • one year ago
  9. calculusfunctions
    Best Response
    You've already chosen the best response.
    Medals 0

    No, I am a math teacher and I will tell you that you must use the quotient rule to find the second derivative, but it would be wise to use the chain rule (power of a function rule) to find the first derivative.

    • one year ago
  10. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    use the quotient rule

    • one year ago
  11. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    @mark_o. Yes, I know. I'm not talking about how to find it, finding it is simple, I'm referring to taking this twice or thrice without having an utter mess of calculations, or without it taking more than some short time. I don't know if there even *is* such a way, I'm just guessing.

    • one year ago
  12. calculusfunctions
    Best Response
    You've already chosen the best response.
    Medals 0

    @mark_o. please stop giving out answers. You're not helping!

    • one year ago
  13. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    I already have the answer. That's not what I'm looking for. I'm wondering if there is some way to manage this in a quick and nice manner, rather than crunching numbers.

    • one year ago
  14. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    (Again, there might *not* be, but I'm not the most imaginative person in Calculus--I deal with Number Theory--and this question came up)

    • one year ago
  15. calculusfunctions
    Best Response
    You've already chosen the best response.
    Medals 0

    Those derivative rules are in place for a reason. You're looking for a short cut and I'm telling you that those derivative rules are the shortcut.

    • one year ago
  16. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    im sorry i didnt want to give the answer on the second derivative., i thought he didnt knoe how to arrive at the first derivative.... continue plzz

    • one year ago
  17. calculusfunctions
    Best Response
    You've already chosen the best response.
    Medals 0

    @mark_o. That's alright! He said he already got the answer so no harm done.

    • one year ago
  18. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, I am not referring to the derivative rules. Of course, most people could come up with the proofs, and I am not looking for a shortcut, at least, not directly. The point that I ask is *not* how to find them, nor to get a lesson on the shortcuts, and whether they should be used or not, I'm asking if there *is* a nice way of *handling* the operations, or if there is some much nicer (not necessarily *easier*) way of dealing with problems that are similar.

    • one year ago
  19. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    ok continue plzz...

    • one year ago
  20. calculusfunctions
    Best Response
    You've already chosen the best response.
    Medals 0

    @mark_o. there is nothing to continue LOL because @LolWolf the answer to your question is a very nice NO.

    • one year ago
  21. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    That's all I wanted to know. Thank you.

    • one year ago
  22. calculusfunctions
    Best Response
    You've already chosen the best response.
    Medals 0

    Welcome!

    • one year ago
  23. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    well, there no way of getting around it,or to short cut them, just continue solving them wether you started on quotient rule or product rule.you will arrive on the same answer,......

    • one year ago
  24. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, thank you.

    • one year ago
  25. mark_o.
    Best Response
    You've already chosen the best response.
    Medals 0

    ok YW, have fun now

    • one year ago
  26. LolWolf
    Best Response
    You've already chosen the best response.
    Medals 0

    Actually, there IS, indeed a much nicer way to do this derivative, if we split it into separate functions. Not to beat a dead horse, or anything, but please don't say that one 'cannot' do something without actually knowing so, even if you're a teacher. If we simply assume the function of \(x^2=l\), then, by repeated application of the chain rule, we reduce the laborious task of handling the repeated quotient rule to a simple product rule with a single multiplication per step, along with a much easier-to-compute chain rule. Enjoy.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.