Derivative of piecewise function with $sinfrac{1}{x}$ term












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I was going through my calculus book, and I am not sure I understand this part



$f(x) = begin{cases} frac{x^2}{4}+x^4sin(frac{1}{x}) &text{if $xneq0$ } \ 0 &text{if $x=0$ } end{cases}$



$ f'(x) = begin{cases} frac{x}{2}-x^2cos(frac{1}{x})+4x^3sin(frac{1}{x}) &text{if $xneq0$ } \ 0 &text{if $x=0$ } end{cases}$



$ f''(x) = begin{cases} frac{1}{2}+12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x}) &text{if $xneq0$ } \ frac{1}{2} &text{if $x=0$ } end{cases}$



So, I know that when I have piecewise function, I need to look at left and right limit, but I don't see why second part in second derivative is $frac{1}{2}$, or rather why does the $sinfrac{1}{x}$ term go to 0?






calculus derivatives







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  • 1




    $begingroup$
    It should be noted that the $sin frac{1}{x}$ term does not "go to $0$", and the result shows that while the second derivative $f"(0)$ exists, the second derivative is not continuous at $x=0$.
    $endgroup$
    – hardmath
    Dec 15 '18 at 4:06
















1












$begingroup$


I was going through my calculus book, and I am not sure I understand this part



$f(x) = begin{cases} frac{x^2}{4}+x^4sin(frac{1}{x}) &text{if $xneq0$ } \ 0 &text{if $x=0$ } end{cases}$



$ f'(x) = begin{cases} frac{x}{2}-x^2cos(frac{1}{x})+4x^3sin(frac{1}{x}) &text{if $xneq0$ } \ 0 &text{if $x=0$ } end{cases}$



$ f''(x) = begin{cases} frac{1}{2}+12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x}) &text{if $xneq0$ } \ frac{1}{2} &text{if $x=0$ } end{cases}$



So, I know that when I have piecewise function, I need to look at left and right limit, but I don't see why second part in second derivative is $frac{1}{2}$, or rather why does the $sinfrac{1}{x}$ term go to 0?






calculus derivatives







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It should be noted that the $sin frac{1}{x}$ term does not "go to $0$", and the result shows that while the second derivative $f"(0)$ exists, the second derivative is not continuous at $x=0$.
    $endgroup$
    – hardmath
    Dec 15 '18 at 4:06














1












1








1


1



$begingroup$


I was going through my calculus book, and I am not sure I understand this part



$f(x) = begin{cases} frac{x^2}{4}+x^4sin(frac{1}{x}) &text{if $xneq0$ } \ 0 &text{if $x=0$ } end{cases}$



$ f'(x) = begin{cases} frac{x}{2}-x^2cos(frac{1}{x})+4x^3sin(frac{1}{x}) &text{if $xneq0$ } \ 0 &text{if $x=0$ } end{cases}$



$ f''(x) = begin{cases} frac{1}{2}+12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x}) &text{if $xneq0$ } \ frac{1}{2} &text{if $x=0$ } end{cases}$



So, I know that when I have piecewise function, I need to look at left and right limit, but I don't see why second part in second derivative is $frac{1}{2}$, or rather why does the $sinfrac{1}{x}$ term go to 0?






calculus derivatives







share|cite|improve this question











$endgroup$




I was going through my calculus book, and I am not sure I understand this part



$f(x) = begin{cases} frac{x^2}{4}+x^4sin(frac{1}{x}) &text{if $xneq0$ } \ 0 &text{if $x=0$ } end{cases}$



$ f'(x) = begin{cases} frac{x}{2}-x^2cos(frac{1}{x})+4x^3sin(frac{1}{x}) &text{if $xneq0$ } \ 0 &text{if $x=0$ } end{cases}$



$ f''(x) = begin{cases} frac{1}{2}+12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x}) &text{if $xneq0$ } \ frac{1}{2} &text{if $x=0$ } end{cases}$



So, I know that when I have piecewise function, I need to look at left and right limit, but I don't see why second part in second derivative is $frac{1}{2}$, or rather why does the $sinfrac{1}{x}$ term go to 0?






calculus derivatives




calculus derivatives






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edited Dec 14 '18 at 19:12









Bernard

121k740116




121k740116










asked Dec 14 '18 at 19:08









DovlaDovla

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  • 1




    $begingroup$
    It should be noted that the $sin frac{1}{x}$ term does not "go to $0$", and the result shows that while the second derivative $f"(0)$ exists, the second derivative is not continuous at $x=0$.
    $endgroup$
    – hardmath
    Dec 15 '18 at 4:06














  • 1




    $begingroup$
    It should be noted that the $sin frac{1}{x}$ term does not "go to $0$", and the result shows that while the second derivative $f"(0)$ exists, the second derivative is not continuous at $x=0$.
    $endgroup$
    – hardmath
    Dec 15 '18 at 4:06








1




1




$begingroup$
It should be noted that the $sin frac{1}{x}$ term does not "go to $0$", and the result shows that while the second derivative $f"(0)$ exists, the second derivative is not continuous at $x=0$.
$endgroup$
– hardmath
Dec 15 '18 at 4:06




$begingroup$
It should be noted that the $sin frac{1}{x}$ term does not "go to $0$", and the result shows that while the second derivative $f"(0)$ exists, the second derivative is not continuous at $x=0$.
$endgroup$
– hardmath
Dec 15 '18 at 4:06










2 Answers
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If $xneq0$, thenbegin{align}frac{f'(x)-f'(0)}x&=frac{frac x2-x^2cosleft(frac1xright)+4x^3sinleft(frac1xright)}x\&=frac12-xcosleft(frac1xright)+4x^2sinleft(frac1xright)end{align}and thereforebegin{align}f''(0)&=lim_{xto0}frac12-xcosleft(frac1xright)+4x^2sinleft(frac1xright)\&=frac12.end{align}






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    3












    $begingroup$

    hint



    $$f''(0)=lim_{xto 0,xne 0}frac{f'(x)-f'(0)}{x-0}$$






    share|cite|improve this answer











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    • $begingroup$
      @hardmath Done. thank you.
      $endgroup$
      – hamam_Abdallah
      Dec 14 '18 at 20:47











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    2 Answers
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    2 Answers
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    5












    $begingroup$

    If $xneq0$, thenbegin{align}frac{f'(x)-f'(0)}x&=frac{frac x2-x^2cosleft(frac1xright)+4x^3sinleft(frac1xright)}x\&=frac12-xcosleft(frac1xright)+4x^2sinleft(frac1xright)end{align}and thereforebegin{align}f''(0)&=lim_{xto0}frac12-xcosleft(frac1xright)+4x^2sinleft(frac1xright)\&=frac12.end{align}






    share|cite|improve this answer









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      5












      $begingroup$

      If $xneq0$, thenbegin{align}frac{f'(x)-f'(0)}x&=frac{frac x2-x^2cosleft(frac1xright)+4x^3sinleft(frac1xright)}x\&=frac12-xcosleft(frac1xright)+4x^2sinleft(frac1xright)end{align}and thereforebegin{align}f''(0)&=lim_{xto0}frac12-xcosleft(frac1xright)+4x^2sinleft(frac1xright)\&=frac12.end{align}






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        If $xneq0$, thenbegin{align}frac{f'(x)-f'(0)}x&=frac{frac x2-x^2cosleft(frac1xright)+4x^3sinleft(frac1xright)}x\&=frac12-xcosleft(frac1xright)+4x^2sinleft(frac1xright)end{align}and thereforebegin{align}f''(0)&=lim_{xto0}frac12-xcosleft(frac1xright)+4x^2sinleft(frac1xright)\&=frac12.end{align}






        share|cite|improve this answer









        $endgroup$



        If $xneq0$, thenbegin{align}frac{f'(x)-f'(0)}x&=frac{frac x2-x^2cosleft(frac1xright)+4x^3sinleft(frac1xright)}x\&=frac12-xcosleft(frac1xright)+4x^2sinleft(frac1xright)end{align}and thereforebegin{align}f''(0)&=lim_{xto0}frac12-xcosleft(frac1xright)+4x^2sinleft(frac1xright)\&=frac12.end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 19:15









        José Carlos SantosJosé Carlos Santos

        160k22127232




        160k22127232























            3












            $begingroup$

            hint



            $$f''(0)=lim_{xto 0,xne 0}frac{f'(x)-f'(0)}{x-0}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @hardmath Done. thank you.
              $endgroup$
              – hamam_Abdallah
              Dec 14 '18 at 20:47
















            3












            $begingroup$

            hint



            $$f''(0)=lim_{xto 0,xne 0}frac{f'(x)-f'(0)}{x-0}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @hardmath Done. thank you.
              $endgroup$
              – hamam_Abdallah
              Dec 14 '18 at 20:47














            3












            3








            3





            $begingroup$

            hint



            $$f''(0)=lim_{xto 0,xne 0}frac{f'(x)-f'(0)}{x-0}$$






            share|cite|improve this answer











            $endgroup$



            hint



            $$f''(0)=lim_{xto 0,xne 0}frac{f'(x)-f'(0)}{x-0}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 14 '18 at 20:47

























            answered Dec 14 '18 at 19:14









            hamam_Abdallahhamam_Abdallah

            38.1k21634




            38.1k21634












            • $begingroup$
              @hardmath Done. thank you.
              $endgroup$
              – hamam_Abdallah
              Dec 14 '18 at 20:47


















            • $begingroup$
              @hardmath Done. thank you.
              $endgroup$
              – hamam_Abdallah
              Dec 14 '18 at 20:47
















            $begingroup$
            @hardmath Done. thank you.
            $endgroup$
            – hamam_Abdallah
            Dec 14 '18 at 20:47




            $begingroup$
            @hardmath Done. thank you.
            $endgroup$
            – hamam_Abdallah
            Dec 14 '18 at 20:47


















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