1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the flaw in this logic?

  1. Feb 5, 2007 #1
    [tex] x^2=x+x+ ... +x [/tex] (with x terms). Its derivative is [tex] 1+1+ ... +1 [/tex] (also x terms). By this logic, [tex] \frac{d}{dx} x^2=x [/tex] But it doesn't. So what's wrong with this picture?
     
  2. jcsd
  3. Feb 5, 2007 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    this is a well known homework problem. So, what makes you think it is valid? What makes you think it isn't?
     
  4. Feb 5, 2007 #3
    I think it's valid because the derivative of the derivative of the sum of any number of functions equals the sum of the derivatives of the functions, and multiplication is just repeated addition.

    I don't think it's valid because when I plug [tex] x^2 [/tex] into the formula [tex] \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex] it comes out as [tex] 2x [/tex].

    I think that it isn't valid, because the book said so, and plugging it into the definition of a derivative is more fundamental (and therefore less likely to be inaccurate) than the rule for the derivative of the sum of functions. However, I can't find the flaw in the first reasoning.
     
    Last edited: Feb 5, 2007
  5. Feb 5, 2007 #4
    You didn't really help much!
     
  6. Feb 5, 2007 #5
    You need to think about when you can apply the rule for derivatives of sums.
     
  7. Feb 5, 2007 #6
    Does this work for every x? What if x is pi? -1?
     
  8. Feb 6, 2007 #7

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    In my mind this is homework. People only get homework help if they actually state what they've done in an attempt. Examine the first premise: that x^2 is x added up x times. That is clearly nonsense.
     
  9. Feb 6, 2007 #8

    Gib Z

    User Avatar
    Homework Helper

    What you just stated works only for natural values of x. The derivative requires a function that is continuous, not descrete over integers.
     
  10. Feb 6, 2007 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You didn't think much. Matt Grime's purpose was to give you something to think about- that's always a great help.

    You might also think about how x+ x+ x (exactly 3 times) is different from (x+ x+ ...+ x) (x times). Each is a function of x in what way?
     
  11. Feb 6, 2007 #10
    Listen: I have thought about it, contrary to what people may think! If x is [tex] \pi [/tex], for example, you just add [tex] \pi [/tex] three times and then add [tex] (\pi - 3) \ast \pi [/tex] and you get [tex] \pi^2 [/tex]. I don't see the problem with applying this method to any real number.
     
    Last edited: Feb 6, 2007
  12. Feb 6, 2007 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That is not the calculation you used before: you said "x2= (x+ x+ ...+ x) (x times)". That is not the same as saying "add [itex]\pi[/itex] to itself 3 times, then add [tex] (\pi - 3) \ast \pi [/tex]." That was d_leet's point. Why do you mean by "[itex]\pi[/itex] added to itself [itex]\pi[/itex] times"?

    My point was not so much that you can't say that "x2= (x+ x+ ...+ x) (x times)" but rather that the derivative of that is not "(1+ 1+ ...+ 1) (x times)". That "(x times)" is another function of x and you haven't "differentiated" that.
     
  13. Feb 6, 2007 #12
    I think the problem itself lies in that there are x terms. I myself find it quite ambiguous but I'm going with it . I'll try and explain:

    Since there are f(x) 'x' terms where f(x) is a function of x. In that we would have to differentiate f(x)*x using product rule which would result: xf'(x)+f(x).
    Thus we cannot directly apply derivatives of sums becuase the number of such sums is in turn a function of x.

    (Well, now! I don't understand that myself!:tongue2: )
     
    Last edited: Feb 6, 2007
  14. Feb 6, 2007 #13

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    that kind of vaguely defined
    Actually pretty well stated.
     
  15. Feb 6, 2007 #14
    LOL, I like to make up stories!
     
  16. Feb 6, 2007 #15
    That makes sense. Thanks a lot. :smile:
     
  17. Feb 6, 2007 #16

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    You gave no indication of that. We aren't psychics

    The 'method' you just used, and the method you gave in the first post are completely different.
     
  18. Feb 6, 2007 #17
    I've already figured out the answer to the question (see my last post), and I don't want to get into any arguments with anyone, so I think this thread should be locked before a fight breaks out or something.
     
  19. Feb 7, 2007 #18

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    What fight? If you read the homework FAQ you will see that homework help is not given unless you show what you've done. That is the primary requirement here for you to get help. So don't complain when you don't get help if you've not done so.
     
  20. Feb 7, 2007 #19

    Gib Z

    User Avatar
    Homework Helper

    Lol Everyone chillax. If you think this is getting heated, just make a new account Izzhov :p CHill, this is math, not girls :)
     
  21. Feb 25, 2007 #20
    It seems you are getting hung up by using the power rule to define the derivative. Power rule is just a short cut for the difference quotient

    lim as h-> 0 [f(x+h)-f(x)]/h.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?