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

asnaseer Group Title

How good is your understanding of differentiation? \(y=x^2\) can be written as \(y=x*x\) which can also be written as \(y=x+x+x+...+x\) (i.e. "x" lots of x). using this we get:\[\begin{align} y&=x^2\\ \therefore \frac{dy}{dx}&=2x\\ &\text{but we can also write this as:}\\ y&=x+x+x+...+x\quad\text{i.e. "x" lots of x}\\ \therefore \frac{dy}{dx}&=1+1+1+…+1\quad\text{i.e. "x" lots of 1's}\\ &=x\\ \therefore 2x&=x\\ \therefore 2&=1 \end{align}\]where is the mistake in this?

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    In the equation y=x+x+x..+x, this is not a continuous function, nor differentiable.

    • 2 years ago
  2. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    it is differentiable and is continuous - just like \(y=x^2\)

    • 2 years ago
  3. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Not really. x here can take only integer values, so it's defined from N to R. While y=x^2 is defined from R to R. So they are tow different functions.

    • 2 years ago
  4. agdgdgdgwngo Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    your mistake is that you didn't differentiate the sum of all the x's properly

    • 2 years ago
  5. agdgdgdgwngo Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    but I forgot the exact mistake :-P

    • 2 years ago
  6. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    you can take one and half lots of something

    • 2 years ago
  7. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    If we put f(x)=y=x+x+..+x (x times), could you tell me what f(0.5) for instance?

    • 2 years ago
  8. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    f(0.5)=0.5+0.5+... (taken 0.5 times) = 0.25

    • 2 years ago
  9. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    how can that be?

    • 2 years ago
  10. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Yeah, how can that be?

    • 2 years ago
  11. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    Mr.Math - you are probably correct in the strictest sense, but agdgdgdgwngo is closer to where the mistake is.

    • 2 years ago
  12. Akshay_Budhkar Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    how is f(0.5)= 0.25??

    • 2 years ago
  13. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    f(x)=x^2=x+x+... (taken x times)

    • 2 years ago
  14. agdgdgdgwngo Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    or \[\sum_{i=1}^{x}x\]

    • 2 years ago
  15. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    I didn't what to use that because it implies x must be an integer

    • 2 years ago
  16. agdgdgdgwngo Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    oh

    • 2 years ago
  17. ktklown Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    What does "taken x times" mean then if not a sum?

    • 2 years ago
  18. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    but I guess Mr.Math is right - we cannot say x^2=x+x+x+... (x times) for non-integers?

    • 2 years ago
  19. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Just quickly 2x=x means x=0, and doesn't mean 2=1.

    • 2 years ago
  20. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    yes Mr.Math - the real question is, how can I get two different results by differentiating the /same/ function?

    • 2 years ago
  21. Akshay_Budhkar Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    x cannot be zero?

    • 2 years ago
  22. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    at ostensibly the same function

    • 2 years ago
  23. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    *at least

    • 2 years ago
  24. agdgdgdgwngo Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    think of the product rule

    • 2 years ago
  25. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't think you can differentiate a variable number of terms like that, but I can't seem to apply the right rule.

    • 2 years ago
  26. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    f(x)=x+x+..+x (x times). The key is "x" times, since we're differentiating with respect to x. If it's f(x)=x+x+..+x ("n" times), what we did would be correct.

    • 2 years ago
  27. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    Mr.Math has hit the nail on the head!

    • 2 years ago
  28. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    but which rule exactly are we breaking and how?

    • 2 years ago
  29. Akshay_Budhkar Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    the product rule?

    • 2 years ago
  30. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    then df/dx=(1+1+1..+1)+(1+1+..+1) (x times) for each brackets I think.

    • 2 years ago
  31. agdgdgdgwngo Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    right

    • 2 years ago
  32. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    if you re-did this using the definition of differentiation then you would get the right result. i.e.\[f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}\]

    • 2 years ago
  33. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    @TuringTest - the being broken is to treat "x" as a constant when it is not

    • 2 years ago
  34. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    *the rule

    • 2 years ago
  35. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Or the chain rule, let x+x+..+x (x times)=u, then we can apply the chain rule I think.

    • 2 years ago
  36. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    that's what I said, a variable number of terms. I just want to make it concrete in my mind.

    • 2 years ago
  37. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    I guess I've also learnt a thing or do from your replies on what a continuous function is - thx all

    • 2 years ago
  38. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    *or two

    • 2 years ago
  39. Akshay_Budhkar Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    but at the first place is x+x+x+... xtimes a continuous function to be differentiated?

    • 2 years ago
  40. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    @Akshay Budhkar - if you use this to evaluate the differential of y = x+x+...+x (x times), then you will get the right answer:\[f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}\]

    • 2 years ago
  41. agdgdgdgwngo Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    btw asnaseer who is the guy in your profile pic? I can't stop wondering! :(

    • 2 years ago
  42. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    that highlights the fact that you cannot treat the x in "x times" as a constant

    • 2 years ago
  43. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    @agdgdgdgwngo - thats me in the pic - why? do I look odd?

    • 2 years ago
  44. Akshay_Budhkar Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    No i was just wondering if the function was continuous, i know that formula, but if a function is not continuous we cannot differentiate it, and i feel this is not a continuous function

    • 2 years ago
  45. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    yes - Mr.Math pointed that out earlier - I guess I used a bad example of trying to illustrate a specific point :-(

    • 2 years ago
  46. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Wait!! \[f'(x)=\lim_{h\to 0}{x+h+x+h+\dots+x+h-x-x-\dots-x \over h}\] \[=\lim_{x\to 0}{xh \over h}=x\]

    • 2 years ago
  47. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    \(h\to 0\)*

    • 2 years ago
  48. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    no Mr.Math - the first set of "x+h" terms will appear "x+h" times

    • 2 years ago
  49. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    This is a old problem.

    • 2 years ago
  50. Akshay_Budhkar Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    you came late ffm that is why it is old :P

    • 2 years ago
  51. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    yes FFM - I came across it several years ago and just recalled it :-)

    • 2 years ago
  52. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    There has been a discussion on this one, let me do some digging.

    • 2 years ago
  53. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    you mean here on OS?

    • 2 years ago
  54. Mr.Math Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Right! \[\lim_{h \to 0}{x+h+x+h+..+x+h-x-x-\dots -x \over h}\] \[=\lim_{h \to 0}{hx+(x+h)h \over h}=2x\]

    • 2 years ago
  55. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    thats it Mr.Math - well done!

    • 2 years ago
  56. Akshay_Budhkar Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    So what is the conclusion of this question ( neglecting the fact that the function is discontinuous) ?

    • 2 years ago
  57. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    the mistake was in treating the "x" in "x times" as a "constant" when differentiating - this is wrong

    • 2 years ago
  58. Akshay_Budhkar Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yea get it :D

    • 2 years ago
  59. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Here is it http://math.stackexchange.com/questions/1096/

    • 2 years ago
  60. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    I wasn't aware of that FFM - I ran into this many many years ago at uni

    • 2 years ago
  61. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    they mention the integer issue as well...

    • 2 years ago
  62. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    It's okay, I can't imagine why you mean many many and many :D

    • 2 years ago
  63. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    ;-D

    • 2 years ago
  64. Akshay_Budhkar Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yea @turing

    • 2 years 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.