Trouble checking whether a function is differentiable or not












0












$begingroup$


I have this function $x^2|cos frac pi x|$ . To check if it's differentiable at $x=0$, I first directly differentiated the given function- having two cases, opening the mod sign with positive and negative signs. In both cases I end up with a $sin frac pi x $ term in the derivative, so that at $x=0$ the derivative is discontinuous (oscillatory discontinuity) and so the derivative shouldn't exist at $x=0$. But, if I differentiate using first principles instead, I get the derivative as $0$, and so the function IS differentiable at $x=0$. So, is the function differentiable or not, which result do I take?






calculus







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  • 2




    $begingroup$
    You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
    $endgroup$
    – metamorphy
    Jan 5 at 13:14








  • 1




    $begingroup$
    To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
    $endgroup$
    – Mark S.
    Jan 5 at 14:05










  • $begingroup$
    But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
    $endgroup$
    – karun mathews
    Jan 5 at 14:43










  • $begingroup$
    Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
    $endgroup$
    – karun mathews
    Jan 5 at 14:48






  • 1




    $begingroup$
    @karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
    $endgroup$
    – Matt Samuel
    Jan 7 at 14:35
















0












$begingroup$


I have this function $x^2|cos frac pi x|$ . To check if it's differentiable at $x=0$, I first directly differentiated the given function- having two cases, opening the mod sign with positive and negative signs. In both cases I end up with a $sin frac pi x $ term in the derivative, so that at $x=0$ the derivative is discontinuous (oscillatory discontinuity) and so the derivative shouldn't exist at $x=0$. But, if I differentiate using first principles instead, I get the derivative as $0$, and so the function IS differentiable at $x=0$. So, is the function differentiable or not, which result do I take?






calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
    $endgroup$
    – metamorphy
    Jan 5 at 13:14








  • 1




    $begingroup$
    To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
    $endgroup$
    – Mark S.
    Jan 5 at 14:05










  • $begingroup$
    But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
    $endgroup$
    – karun mathews
    Jan 5 at 14:43










  • $begingroup$
    Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
    $endgroup$
    – karun mathews
    Jan 5 at 14:48






  • 1




    $begingroup$
    @karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
    $endgroup$
    – Matt Samuel
    Jan 7 at 14:35














0












0








0





$begingroup$


I have this function $x^2|cos frac pi x|$ . To check if it's differentiable at $x=0$, I first directly differentiated the given function- having two cases, opening the mod sign with positive and negative signs. In both cases I end up with a $sin frac pi x $ term in the derivative, so that at $x=0$ the derivative is discontinuous (oscillatory discontinuity) and so the derivative shouldn't exist at $x=0$. But, if I differentiate using first principles instead, I get the derivative as $0$, and so the function IS differentiable at $x=0$. So, is the function differentiable or not, which result do I take?






calculus







share|cite|improve this question











$endgroup$




I have this function $x^2|cos frac pi x|$ . To check if it's differentiable at $x=0$, I first directly differentiated the given function- having two cases, opening the mod sign with positive and negative signs. In both cases I end up with a $sin frac pi x $ term in the derivative, so that at $x=0$ the derivative is discontinuous (oscillatory discontinuity) and so the derivative shouldn't exist at $x=0$. But, if I differentiate using first principles instead, I get the derivative as $0$, and so the function IS differentiable at $x=0$. So, is the function differentiable or not, which result do I take?






calculus




calculus






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 13:13









Kenny Wong

19.3k21441




19.3k21441










asked Jan 5 at 13:09









karun mathewskarun mathews

245




245








  • 2




    $begingroup$
    You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
    $endgroup$
    – metamorphy
    Jan 5 at 13:14








  • 1




    $begingroup$
    To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
    $endgroup$
    – Mark S.
    Jan 5 at 14:05










  • $begingroup$
    But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
    $endgroup$
    – karun mathews
    Jan 5 at 14:43










  • $begingroup$
    Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
    $endgroup$
    – karun mathews
    Jan 5 at 14:48






  • 1




    $begingroup$
    @karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
    $endgroup$
    – Matt Samuel
    Jan 7 at 14:35














  • 2




    $begingroup$
    You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
    $endgroup$
    – metamorphy
    Jan 5 at 13:14








  • 1




    $begingroup$
    To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
    $endgroup$
    – Mark S.
    Jan 5 at 14:05










  • $begingroup$
    But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
    $endgroup$
    – karun mathews
    Jan 5 at 14:43










  • $begingroup$
    Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
    $endgroup$
    – karun mathews
    Jan 5 at 14:48






  • 1




    $begingroup$
    @karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
    $endgroup$
    – Matt Samuel
    Jan 7 at 14:35








2




2




$begingroup$
You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
$endgroup$
– metamorphy
Jan 5 at 13:14






$begingroup$
You've shown that the derivative exists (and in addition that it is discontinuous at $0$). (You're done.)
$endgroup$
– metamorphy
Jan 5 at 13:14






1




1




$begingroup$
To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
$endgroup$
– Mark S.
Jan 5 at 14:05




$begingroup$
To be clear, "differentiable" means "the derivative exists" (which you showed from first principles). It does not mean "the derivative is itself continuous" (which you showed is false here).
$endgroup$
– Mark S.
Jan 5 at 14:05












$begingroup$
But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
$endgroup$
– karun mathews
Jan 5 at 14:43




$begingroup$
But I don't understand, if by differentiation using first principles, I could find the left hand and right hand limits of the derivative, then why is the function still discontinuous? Is it because we can't find the value of the derivative AT x=0?
$endgroup$
– karun mathews
Jan 5 at 14:43












$begingroup$
Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
$endgroup$
– karun mathews
Jan 5 at 14:48




$begingroup$
Also, why would I not get a left hand and right hand limit by differentiating directly, how is that process wrong?
$endgroup$
– karun mathews
Jan 5 at 14:48




1




1




$begingroup$
@karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
$endgroup$
– Matt Samuel
Jan 7 at 14:35




$begingroup$
@karun Yes. A function of this type is a standard example. Another is $x^2sin frac1x$.
$endgroup$
– Matt Samuel
Jan 7 at 14:35










2 Answers
2






active

oldest

votes


















0












$begingroup$

Hint: Compute $$lim_{hto 0}frac{f(0+h)-f(0)}{h}$$ this is $$lim_{hto 0}frac{h^2|cos(frac{pi}{h})|-0}{h}$$ and since $$frac{h^2|cos(frac{pi}{h})|}{h}=h|cos(frac{pi}{h})|le h$$ this tends to zero for $h$ tends to zero.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    He has already computed this.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:14










  • $begingroup$
    Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
    $endgroup$
    – karun mathews
    Jan 5 at 13:55










  • $begingroup$
    @karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
    $endgroup$
    – Matt Samuel
    Jan 5 at 17:30



















0












$begingroup$

Apparently this type of function is an example of a function whose derivative at a point exists (this can be checked by applying the first principle of differentiation), however it's derivative is also discontinuous at that very same point. Such functions do exist. Another example of this is $x^2sin1/x$ .
Thanks @Matt Samuel for explaining






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Hint: Compute $$lim_{hto 0}frac{f(0+h)-f(0)}{h}$$ this is $$lim_{hto 0}frac{h^2|cos(frac{pi}{h})|-0}{h}$$ and since $$frac{h^2|cos(frac{pi}{h})|}{h}=h|cos(frac{pi}{h})|le h$$ this tends to zero for $h$ tends to zero.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      He has already computed this.
      $endgroup$
      – Thomas Shelby
      Jan 5 at 13:14










    • $begingroup$
      Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
      $endgroup$
      – karun mathews
      Jan 5 at 13:55










    • $begingroup$
      @karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
      $endgroup$
      – Matt Samuel
      Jan 5 at 17:30
















    0












    $begingroup$

    Hint: Compute $$lim_{hto 0}frac{f(0+h)-f(0)}{h}$$ this is $$lim_{hto 0}frac{h^2|cos(frac{pi}{h})|-0}{h}$$ and since $$frac{h^2|cos(frac{pi}{h})|}{h}=h|cos(frac{pi}{h})|le h$$ this tends to zero for $h$ tends to zero.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      He has already computed this.
      $endgroup$
      – Thomas Shelby
      Jan 5 at 13:14










    • $begingroup$
      Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
      $endgroup$
      – karun mathews
      Jan 5 at 13:55










    • $begingroup$
      @karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
      $endgroup$
      – Matt Samuel
      Jan 5 at 17:30














    0












    0








    0





    $begingroup$

    Hint: Compute $$lim_{hto 0}frac{f(0+h)-f(0)}{h}$$ this is $$lim_{hto 0}frac{h^2|cos(frac{pi}{h})|-0}{h}$$ and since $$frac{h^2|cos(frac{pi}{h})|}{h}=h|cos(frac{pi}{h})|le h$$ this tends to zero for $h$ tends to zero.






    share|cite|improve this answer











    $endgroup$



    Hint: Compute $$lim_{hto 0}frac{f(0+h)-f(0)}{h}$$ this is $$lim_{hto 0}frac{h^2|cos(frac{pi}{h})|-0}{h}$$ and since $$frac{h^2|cos(frac{pi}{h})|}{h}=h|cos(frac{pi}{h})|le h$$ this tends to zero for $h$ tends to zero.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 5 at 13:16

























    answered Jan 5 at 13:13









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    78.6k42867




    78.6k42867












    • $begingroup$
      He has already computed this.
      $endgroup$
      – Thomas Shelby
      Jan 5 at 13:14










    • $begingroup$
      Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
      $endgroup$
      – karun mathews
      Jan 5 at 13:55










    • $begingroup$
      @karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
      $endgroup$
      – Matt Samuel
      Jan 5 at 17:30


















    • $begingroup$
      He has already computed this.
      $endgroup$
      – Thomas Shelby
      Jan 5 at 13:14










    • $begingroup$
      Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
      $endgroup$
      – karun mathews
      Jan 5 at 13:55










    • $begingroup$
      @karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
      $endgroup$
      – Matt Samuel
      Jan 5 at 17:30
















    $begingroup$
    He has already computed this.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:14




    $begingroup$
    He has already computed this.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:14












    $begingroup$
    Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
    $endgroup$
    – karun mathews
    Jan 5 at 13:55




    $begingroup$
    Ya, I already did this- that's what I meant by differentiation by first principles, sorry I didn't elaborate. My problem is that when I directly differentiate, I get that the derivative doesn't exist, because of that sin(π/x) term- which will become sin(infinity) at x=0.
    $endgroup$
    – karun mathews
    Jan 5 at 13:55












    $begingroup$
    @karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
    $endgroup$
    – Matt Samuel
    Jan 5 at 17:30




    $begingroup$
    @karun There's no such thing as $sin(infty)$. All you know is that $xto 0$, $sin(fracpi{x})$ is bounded in absolute value by $1$. This is good enough to show the limit is $0$.
    $endgroup$
    – Matt Samuel
    Jan 5 at 17:30











    0












    $begingroup$

    Apparently this type of function is an example of a function whose derivative at a point exists (this can be checked by applying the first principle of differentiation), however it's derivative is also discontinuous at that very same point. Such functions do exist. Another example of this is $x^2sin1/x$ .
    Thanks @Matt Samuel for explaining






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Apparently this type of function is an example of a function whose derivative at a point exists (this can be checked by applying the first principle of differentiation), however it's derivative is also discontinuous at that very same point. Such functions do exist. Another example of this is $x^2sin1/x$ .
      Thanks @Matt Samuel for explaining






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Apparently this type of function is an example of a function whose derivative at a point exists (this can be checked by applying the first principle of differentiation), however it's derivative is also discontinuous at that very same point. Such functions do exist. Another example of this is $x^2sin1/x$ .
        Thanks @Matt Samuel for explaining






        share|cite|improve this answer









        $endgroup$



        Apparently this type of function is an example of a function whose derivative at a point exists (this can be checked by applying the first principle of differentiation), however it's derivative is also discontinuous at that very same point. Such functions do exist. Another example of this is $x^2sin1/x$ .
        Thanks @Matt Samuel for explaining







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 19 at 11:00









        karun mathewskarun mathews

        245




        245






























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