Derivative of $sqrt{x^{2}}$ at $x=0$












3












$begingroup$


I'm suppose to calculate the derivative of $f(x)=sqrt{x^2}$ when $x=0$.
I.e., I need to determine $f'(0)$. I worked it out this way:



$begin{align} f'(0 )&= lim_{xrightarrow 0} frac{f(x)-f(a)}{x-a}\ \
&=lim_{xrightarrow 0} frac{sqrt{x^2} - 0}{x-0}\\
&=lim_{xrightarrow 0} frac{x}{x}\ \
&=1end{align}$



I know I'm doing something wrong, because the solution says there is no derivative, But I don't know why.






calculus derivatives







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    3












    $begingroup$


    I'm suppose to calculate the derivative of $f(x)=sqrt{x^2}$ when $x=0$.
    I.e., I need to determine $f'(0)$. I worked it out this way:



    $begin{align} f'(0 )&= lim_{xrightarrow 0} frac{f(x)-f(a)}{x-a}\ \
    &=lim_{xrightarrow 0} frac{sqrt{x^2} - 0}{x-0}\\
    &=lim_{xrightarrow 0} frac{x}{x}\ \
    &=1end{align}$



    I know I'm doing something wrong, because the solution says there is no derivative, But I don't know why.






    calculus derivatives







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I'm suppose to calculate the derivative of $f(x)=sqrt{x^2}$ when $x=0$.
      I.e., I need to determine $f'(0)$. I worked it out this way:



      $begin{align} f'(0 )&= lim_{xrightarrow 0} frac{f(x)-f(a)}{x-a}\ \
      &=lim_{xrightarrow 0} frac{sqrt{x^2} - 0}{x-0}\\
      &=lim_{xrightarrow 0} frac{x}{x}\ \
      &=1end{align}$



      I know I'm doing something wrong, because the solution says there is no derivative, But I don't know why.






      calculus derivatives







      share|cite|improve this question











      $endgroup$




      I'm suppose to calculate the derivative of $f(x)=sqrt{x^2}$ when $x=0$.
      I.e., I need to determine $f'(0)$. I worked it out this way:



      $begin{align} f'(0 )&= lim_{xrightarrow 0} frac{f(x)-f(a)}{x-a}\ \
      &=lim_{xrightarrow 0} frac{sqrt{x^2} - 0}{x-0}\\
      &=lim_{xrightarrow 0} frac{x}{x}\ \
      &=1end{align}$



      I know I'm doing something wrong, because the solution says there is no derivative, But I don't know why.






      calculus derivatives




      calculus derivatives






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 17 at 19:01









      jordan_glen

      1




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      asked Jan 17 at 18:22









      JakcjonesJakcjones

      828




      828






















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          $begingroup$

          Your issue is the following: $sqrt{x^{2}}=|x|$, not $x$. So your problem reduces to
          $$lim_{xto 0}frac{|x|}{x}$$
          which does not exist. This is because the right-hand limit ($1$) and left-hand limit $(-1)$ do not agree.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Ok, it makes sense. So in order for the derivative to exist, both sides of the limit have to be equal. Got it. Ty
            $endgroup$
            – Jakcjones
            Jan 17 at 18:27













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          1






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          active

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          9












          $begingroup$

          Your issue is the following: $sqrt{x^{2}}=|x|$, not $x$. So your problem reduces to
          $$lim_{xto 0}frac{|x|}{x}$$
          which does not exist. This is because the right-hand limit ($1$) and left-hand limit $(-1)$ do not agree.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Ok, it makes sense. So in order for the derivative to exist, both sides of the limit have to be equal. Got it. Ty
            $endgroup$
            – Jakcjones
            Jan 17 at 18:27


















          9












          $begingroup$

          Your issue is the following: $sqrt{x^{2}}=|x|$, not $x$. So your problem reduces to
          $$lim_{xto 0}frac{|x|}{x}$$
          which does not exist. This is because the right-hand limit ($1$) and left-hand limit $(-1)$ do not agree.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Ok, it makes sense. So in order for the derivative to exist, both sides of the limit have to be equal. Got it. Ty
            $endgroup$
            – Jakcjones
            Jan 17 at 18:27
















          9












          9








          9





          $begingroup$

          Your issue is the following: $sqrt{x^{2}}=|x|$, not $x$. So your problem reduces to
          $$lim_{xto 0}frac{|x|}{x}$$
          which does not exist. This is because the right-hand limit ($1$) and left-hand limit $(-1)$ do not agree.






          share|cite|improve this answer









          $endgroup$



          Your issue is the following: $sqrt{x^{2}}=|x|$, not $x$. So your problem reduces to
          $$lim_{xto 0}frac{|x|}{x}$$
          which does not exist. This is because the right-hand limit ($1$) and left-hand limit $(-1)$ do not agree.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 17 at 18:24









          pwerthpwerth

          3,243417




          3,243417








          • 2




            $begingroup$
            Ok, it makes sense. So in order for the derivative to exist, both sides of the limit have to be equal. Got it. Ty
            $endgroup$
            – Jakcjones
            Jan 17 at 18:27
















          • 2




            $begingroup$
            Ok, it makes sense. So in order for the derivative to exist, both sides of the limit have to be equal. Got it. Ty
            $endgroup$
            – Jakcjones
            Jan 17 at 18:27










          2




          2




          $begingroup$
          Ok, it makes sense. So in order for the derivative to exist, both sides of the limit have to be equal. Got it. Ty
          $endgroup$
          – Jakcjones
          Jan 17 at 18:27






          $begingroup$
          Ok, it makes sense. So in order for the derivative to exist, both sides of the limit have to be equal. Got it. Ty
          $endgroup$
          – Jakcjones
          Jan 17 at 18:27




















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