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

frx Group Title

Find, by induction, the n:th derivitive of \[\frac{ x ^{2}+2x-3 }{ 4x+4 }\] I've counted the 4 first derivitives which are: 1. \[\frac{ x ^{2}+2x+5 }{ 4(x+1)^{2} }\] 2. \[\frac{ -2 }{ (x+1)^{3} }\] 3. \[\frac{ 6 }{ (x+1)^{4} }\] 4. \[\frac{ -24 }{ (x+1)^{5} }\] 5. \[\frac{ (-1)^{n+1}*?*n }{ (x+1)^{n+1} } \] I don't really know if i'm on the right track cause i didn't find the n:th derivitive by induction..

  • one year ago
  • one year ago

  • This Question is Closed
  1. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\text{ Let } f^{(0)}(x)=f(x)\] \[\text{ Let } f^{(i)}(x) \text{ be the ith derivative of } f \] So we have \[f^{(0)}(x)=\frac{x^2+2x-3}{4x+4}=\frac{x^2+2x-3}{4(x+1)}\] \[f^{(0)}(x)=\frac{1}{4} \cdot \frac{x^2+2x-3}{x+1}\] \[f^{(1)}(x)=\frac{1}{4} \cdot \frac{(2x+2)(x+1)-(x^2+2x-3)(1)}{(x+1)^2}\] \[f^{(1)}(x)=\frac{1}{4} \cdot \frac{2x^2+2x+2x+2-x^2-2x+3}{(x+1)^2}\] \[f^{(1)}(x)=\frac{1}{4} \cdot \frac{x^2+2x+5}{(x+1)^2}\]

    • one year ago
  2. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok so good so far...you have the 1st derivative right... Okay...checking your second

    • one year ago
  3. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Great!

    • one year ago
  4. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[f^{(2)}(x)=\frac{1}{4} \frac{(2x+2)(x+1)^2-(x^2+2x+5) \cdot 2(x+1)}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4} \frac{(2x+2)(x^2+2x+1)-(2x^3+2x^2+4x^2+4x+10x+10)}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4} \frac{(2x^3+4x^2+2x+2x^2+4x+2)-(2x^3+6x^2+14x+10)}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4}\frac{(2x^3+6x^2+6x+2)-2x^3-6x^2-14x-10}{(x+1)^4}\] \[f^{(2)}(x)=\frac{1}{4}\frac{-8x-8}{(x+1)^4}=\frac{1}{4} \frac{-8(x+1)}{(x+4)^4}\] \[f^{(2)}(x)=\frac{-8}{4} \frac{1}{(x+4)^3}\]

    • one year ago
  5. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok that one is fine...I'm going to assume your others one are correct then lol

    • one year ago
  6. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Haha, yeah I think they are, a lot of work you're doing ;)

    • one year ago
  7. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok..I see part of it...one sec.

    • one year ago
  8. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[f^{(2)}=\frac{2 \cdot 3}{(x+1)^3} , f^{(3)}= - \frac{2 \cdot 3 \cdot 4}{(x+1)^4}\]

    • one year ago
  9. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    You notice factorial is involved?

    • one year ago
  10. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I just notice the exponents and how the top was changing.

    • one year ago
  11. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, I didn't actually

    • one year ago
  12. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    So you almost had it :)

    • one year ago
  13. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    But the (-1)^n+1 should be there to control the signs right?

    • one year ago
  14. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Right! Totally.

    • one year ago
  15. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Your only expression that doesn't work is n=1 So I would say for your expression n>=2

    • one year ago
  16. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    That is when the pattern starts to occur

    • one year ago
  17. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    And you want for n=2 for the expression to be negative so yeah n+1 works (-1)^{n+1} is -1 for n=2 :)

    • one year ago
  18. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    So the it's \[\frac{ (-1)^{n+1} n! }{ (x+1)^{n+1} }\] \[n\]

    • one year ago
  19. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    n>=2

    • one year ago
  20. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yep that is exactly right :)

    • one year ago
  21. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Wonderful! But what really made me confused was the induction think, so is the task now to prove this for n>=2 by induction or are we done?

    • one year ago
  22. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    thing*

    • one year ago
  23. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Find by induction it says, so isn't this the wrong way?

    • one year ago
  24. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I think it is weird it says find by induction.

    • one year ago
  25. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Usually I see show or prove by induction.

    • one year ago
  26. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    And I know what that means when it is written but find by induction I have trouble understanding what they mean

    • one year ago
  27. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok, maybe they just mean proove the n:th derivitive by induction

    • one year ago
  28. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    If that is the case, I can help you with that

    • one year ago
  29. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I assume that's it. We've never talked about finding the derivitive by induction anyway

    • one year ago
  30. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    So I will not be too fancy in helping you with this proof... I will just give you what you need and you can make it all fancy if you choose So you need to show it is true for n=2 Then assume it is true for some integer n=k That is, that we are assuming \[f^{(k)}(x)=\frac{(-1)^{k+1} k!}{(x+1)^{k+1}}\] Now you want to show that it is true for n=k+1 That is, you want to show that \[f^{(k+1)}(x)=\frac{(-1)^{(k+1)+1}(k+1)!}{(x+1)^{(k+1)+1}}\]

    • one year ago
  31. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Recall, \[f^{(k+1)}(x)=(f^{(k)}(x))'\]

    • one year ago
  32. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    If you want it might make it easier if you do an odd k version and an even k version I think I would do that I see nothing wrong with it

    • one year ago
  33. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, so proving k+1 is just the derivitive of n=k ?

    • one year ago
  34. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    That makes perfect sense

    • one year ago
  35. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yep just like \[f^{(3)}(x)=(f^{(2)}(x))'\]

    • one year ago
  36. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Or for any known value k

    • one year ago
  37. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Exactly, I get it!

    • one year ago
  38. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    A million thank you myininaya!

    • one year ago
  39. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Really great tutor!

    • one year ago
  40. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Aww...Thanks. But let me know if you run into any trouble with the derivative part Don't forget k is just a constant So don't look to scared when differentiating

    • one year ago
  41. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    too*

    • one year ago
  42. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    One last question, the derivitive of the factorial part, how does that work?

    • one year ago
  43. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ohh I see it know, it nothing but a constant to :)

    • one year ago
  44. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    now*

    • one year ago
  45. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    You know what I don't think you actually need to do an even k and odd k

    • one year ago
  46. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    It doesn't matter if k is even or odd, you can still do it as one case :)

    • one year ago
  47. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah, it works out nicely :)

    • one year ago
  48. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Awesome! I tag you later on if i get any trouble but i doubt it :) Once again, thank you so much!:D

    • one year ago
  49. myininaya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    No problem. I like your questions. I think I remember helping you with some other problems you presented.

    • one year ago
  50. frx Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    You did, really valueble!

    • 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.