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Derivative of function only using definition?

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]f(x) = \left\{ {\begin{array}{*{20}{c}}
    {{x^2}\sin \frac{1}{x}}&{x \ne 0}\\
    0&{x = 0}
    \end{array}} \right.[/tex]

    Is it differentiable at x=0? If it is, what's its value?


    2. Relevant equations



    3. The attempt at a solution
    I've calculated the derivative function for x not equal zero:
    [tex]f'(x) = \left\{ {\begin{array}{*{20}{c}}
    {2x\sin \frac{1}{x} - \cos \frac{1}{x}}&{x \ne 0}\\
    0&{x = 0}
    \end{array}} \right.[/tex]

    And:
    [tex]\mathop {\lim }\limits_{x \to 0} 2x\sin \frac{1}{x} - \cos \frac{1}{x}[/tex]

    This limit doesn't exists so IT IS NOT DIFFERENTIABLE at 0. But if I use the definition of the incremental I DO get THAT IT IS differentiable at 0 and the derivative is 0. How could this be possible?


    Please, try to be the most clear as you can, I get easily confused with this things... Thanks!
     
    Last edited: Apr 13, 2012
  2. jcsd
  3. Apr 13, 2012 #2
    Yes, you've shown that f' is not continuous, but that's all that you've shown! In general, a function may be continuous, may even be differentiable, but then the derivative of the function need not be continuous or differentiable. This can happen fairly easily, in fact, when you go defining functions piece-wise.
     
  4. Apr 13, 2012 #3
    I really appreciate your help but your reply is extremelly poor and doesn't help me at all.
     
  5. Apr 13, 2012 #4
    The question is whether f' exists at 0, not whether f' is continuous at 0. It's probably best to use the definition of derivative when the function is defined piece-wise like this.
     
  6. Apr 13, 2012 #5
    Well, yeah, I knew that already, it's not only "the best", it's the only one way to know if it differentiable but you're not explaining anything about it. Why did I get it was not differentiable at 0 when I did the limit of f'(x) at 0? What's the mistake? If f' is not cont. how could it be differentiable? Seriouosly, I really really appreciate your attempt at helping me but it's useless if you just reply 2 lines saying nothing new, the only thing you do is to confuse me even more and don't aport anything.
     
  7. Apr 13, 2012 #6

    Dick

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    The definition of derivative everybody is talking about is limit h->0 (f(0+h)-f(0))/h.
     
  8. Apr 13, 2012 #7
    And? does it aport something I don't know already? I think you're not understanding what my question is. Re-read the principal message, I've edited it.

    Thanks!
     
  9. Apr 13, 2012 #8

    Dick

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    Ok, yes, I see you've already done that. So f'(0) is 0 and it IS differentiable at 0. But f'(x) doesn't have a limit as x->0. There's nothing really wrong with that. It just means f'(x) IS NOT continuous at x=0. It is possible, you've just worked through the example. Why do you think it isn't? f(x) is differentiable everywhere, but doesn't have continuous limits.
     
  10. Apr 13, 2012 #9

    Fredrik

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    What sort of answer do you want? The fact that f' is not continuous implies that f' is not differentiable. Maybe it would be easier to figure out what to tell you if you explained why that bothers you. Did you think that "f' is not differentiable" implies that "f is not differentiable"?

    You probably shouldn't be telling people who are trying to help you that their answers are "extremely poor", especially when they're not.
     
  11. Apr 13, 2012 #10
    I'm going to tell my understanding (there's surely sth wrong):
    when we calculate derivatives at point a we just use the product rules, sum, etc. and in this case I don't understand why is not possible to do that because if we do so we get that it's not differentiable at 0 given the fact that:


    [tex]\mathop {\lim }\limits_{x \to 0} 2x\sin \frac{1}{x} - \cos \frac{1}{x}[/tex] Doesn't exist.

    So why can't we apply this but the definition?
     
    Last edited: Apr 13, 2012
  12. Apr 13, 2012 #11
    You cannot say whether the replies are helping me or not becuase you're merely not me. I don't want to be rude, I really appreciate their help. But maybe there're considering many things obvious that are not for me, so there's almost zero help.

    In this case f' is not continuous at 0 but f is differentiable at 0. Apart from that I got it's differentiable and not differentiable at the same point. That's what I don't get.
     
  13. Apr 13, 2012 #12

    Dick

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    The form x^2*sin(1/x) is NOT VALID at x=0. It's not defined at x=0. You can't use it at x=0. Being differentiable doesn't mean you can take the formula for the derivative where it is valid and take a limit to get it at another point. The function is differentiable. Your LIMIT operation is incorrect.
     
  14. Apr 13, 2012 #13
    Mmmm... so
    1. if f is continuous, f' doesn't necesarilly have to be continuous?

    2. If we know f it's continuous at 0 then we could take the derivative at 0 which is 0:

    [tex]f'(x) = \left\{ {\begin{array}{*{20}{c}}
    {2x\sin \frac{1}{x} - \cos \frac{1}{x}}&{x \ne 0}\\
    0&{x = 0}
    \end{array}} \right.[/tex]

    And there's no need to use the definition, am I right?

    Thanks!!
     
  15. Apr 13, 2012 #14

    Fredrik

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    Maybe I misunderstood your argument. It seemed to me that you were saying that you had found that f is differentiable and f' is not continuous. I don't see why that would be a problem. The result that f is differentiable implies that f is continuous, not that f' is continuous. Similarly, the result that f' is not continuous implies that f' is not differentiable, not that f is not differentiable.

    Edit: I first said that I think you got the wrong result for f'(x) here, but that was actually my mistake. Sorry about that.
     
  16. Apr 13, 2012 #15

    Dick

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    1. Yes, f can be continuous and f' discontinuous. This is an example of that.
    2. HOW are you concluding that f'(x)=0? You DO need to use the difference quotient to prove that!
     
  17. Apr 13, 2012 #16
    Why cannot I say that f'(0)=0? I know that f is continuous at 0 because x^2.sin(1/x) approaches to 0 when x->0. And I know that the derivative function for x=0 is 0. Given that 2 conditions is why I conclude it's differentable.
     
  18. Apr 13, 2012 #17

    Dick

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    Try this. Take the same problem but change x^2*sin(1/x) to x*sin(1/x). Do you still conclude f'(x)=0?
     
  19. Apr 13, 2012 #18
    Yes. But by definition no. What's wrong about my thinking?
     
  20. Apr 13, 2012 #19

    Dick

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    It's a little hard to say. Are you thinking that if I define:

    [tex]f(x) = \left\{ {\begin{array}{*{20}{c}}
    {g(x)}&{x \ne 0}\\
    0&{x = 0}
    \end{array}} \right.[/tex]

    and limit x->0 g(x)=0 then it must be true that f'(0)=0? That's just plain not true. Take g(x)=x.
     
    Last edited: Apr 13, 2012
  21. Apr 13, 2012 #20

    Dick

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    Perhaps you are thinking since f(x) is defined piecewise, then you should find f'(x) by differentiating on each piece and since the piece where x=0 is defined to be the constant 0, f'(0) must be 0?? That would be wrong thinking.
     
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