The antiderivative of the absolute value is apparently x|x| / 2, but when you differentiate that you don't...












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Unless I'm making a mistake (which I probably am), when you differentiate (x * |x|) / 2, you don't get |x|, which means (x * |x|) / 2 isn't an antiderivative of |x|?



Steps for differentiating (x |x|) / 2:



First, note that the derivative of (|x| / 2) is x / (2 |x|). I got this just by using the fact that |x| = (x^2)^1/2 and applying the chain rule.



Split it up: (x * |x|) / 2 = (x / 2) * (|x| / 2)



Apply the product rule: ( 1 * (|x| / 2) ) + ( (x / 2) * (x / 2 |x|) )



= (|x| * 2) + (x^2 / 4|x|)



So I'm getting the derivative as (|x| * 2) + (x^2 / 4|x|) ... where's my mistake? I think I differentiated everything correctly.



Any help is greatly appreciated.






calculus integration derivatives







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    0












    $begingroup$


    Unless I'm making a mistake (which I probably am), when you differentiate (x * |x|) / 2, you don't get |x|, which means (x * |x|) / 2 isn't an antiderivative of |x|?



    Steps for differentiating (x |x|) / 2:



    First, note that the derivative of (|x| / 2) is x / (2 |x|). I got this just by using the fact that |x| = (x^2)^1/2 and applying the chain rule.



    Split it up: (x * |x|) / 2 = (x / 2) * (|x| / 2)



    Apply the product rule: ( 1 * (|x| / 2) ) + ( (x / 2) * (x / 2 |x|) )



    = (|x| * 2) + (x^2 / 4|x|)



    So I'm getting the derivative as (|x| * 2) + (x^2 / 4|x|) ... where's my mistake? I think I differentiated everything correctly.



    Any help is greatly appreciated.






    calculus integration derivatives







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Unless I'm making a mistake (which I probably am), when you differentiate (x * |x|) / 2, you don't get |x|, which means (x * |x|) / 2 isn't an antiderivative of |x|?



      Steps for differentiating (x |x|) / 2:



      First, note that the derivative of (|x| / 2) is x / (2 |x|). I got this just by using the fact that |x| = (x^2)^1/2 and applying the chain rule.



      Split it up: (x * |x|) / 2 = (x / 2) * (|x| / 2)



      Apply the product rule: ( 1 * (|x| / 2) ) + ( (x / 2) * (x / 2 |x|) )



      = (|x| * 2) + (x^2 / 4|x|)



      So I'm getting the derivative as (|x| * 2) + (x^2 / 4|x|) ... where's my mistake? I think I differentiated everything correctly.



      Any help is greatly appreciated.






      calculus integration derivatives







      share|cite|improve this question









      $endgroup$




      Unless I'm making a mistake (which I probably am), when you differentiate (x * |x|) / 2, you don't get |x|, which means (x * |x|) / 2 isn't an antiderivative of |x|?



      Steps for differentiating (x |x|) / 2:



      First, note that the derivative of (|x| / 2) is x / (2 |x|). I got this just by using the fact that |x| = (x^2)^1/2 and applying the chain rule.



      Split it up: (x * |x|) / 2 = (x / 2) * (|x| / 2)



      Apply the product rule: ( 1 * (|x| / 2) ) + ( (x / 2) * (x / 2 |x|) )



      = (|x| * 2) + (x^2 / 4|x|)



      So I'm getting the derivative as (|x| * 2) + (x^2 / 4|x|) ... where's my mistake? I think I differentiated everything correctly.



      Any help is greatly appreciated.






      calculus integration derivatives




      calculus integration derivatives






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      asked Dec 2 '18 at 15:43









      James RonaldJames Ronald

      1057




      1057






















          3 Answers
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          active

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          4












          $begingroup$

          It happens that$$frac{xlvert xrvert}2=xtimesfrac{lvert xrvert}2;$$you got the $frac12$ part in both factors, which is wrong.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Add to that the missing $2$ in the denominator of the first term after applying product rule.
            $endgroup$
            – Shubham Johri
            Dec 2 '18 at 15:57



















          0












          $begingroup$

          Consider what the product of $x$ and $|x|$ should look like. You'd get a piecewise function which is continuous.
          It is $-x^2$ for $x<0$ and $x^2$ for $x>0$.



          Differentiate this along those intervals.
          You should get $2|x|$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hm I'll try this out, thanks!
            $endgroup$
            – James Ronald
            Dec 2 '18 at 16:40



















          0












          $begingroup$

          False statement in your argument:
          $$
          frac{x|x|}{2}=frac{x}{2}frac{|x|}{2}
          $$



          Get rid of the factor $1/2$, but note that you cannot apply the product rule at $0$, because $|x|$ is not differentiable there.



          With the product rule you have, for $xne0$,
          $$
          F'(x)=frac{1}{2}(x|x|)'=frac{1}{2}left(|x|+xfrac{|x|}{x}right)=frac{1}{2}2|x|=|x|
          $$



          Now we also have to compute $F'(0)$:
          $$
          F'(0)=lim_{xto0}frac{F(x)-F(0)}{x}=lim_{xto0}frac{x|x|/2}{x}=
          lim_{xto0}frac{|x|}{2}=0
          $$
          Hence $F'(x)=|x|$ for every $x$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks so much for showing the rest of the steps as well, I understand now. Thank you!
            $endgroup$
            – James Ronald
            Dec 2 '18 at 16:40











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          3 Answers
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          3 Answers
          3






          active

          oldest

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          active

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          active

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          4












          $begingroup$

          It happens that$$frac{xlvert xrvert}2=xtimesfrac{lvert xrvert}2;$$you got the $frac12$ part in both factors, which is wrong.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Add to that the missing $2$ in the denominator of the first term after applying product rule.
            $endgroup$
            – Shubham Johri
            Dec 2 '18 at 15:57
















          4












          $begingroup$

          It happens that$$frac{xlvert xrvert}2=xtimesfrac{lvert xrvert}2;$$you got the $frac12$ part in both factors, which is wrong.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Add to that the missing $2$ in the denominator of the first term after applying product rule.
            $endgroup$
            – Shubham Johri
            Dec 2 '18 at 15:57














          4












          4








          4





          $begingroup$

          It happens that$$frac{xlvert xrvert}2=xtimesfrac{lvert xrvert}2;$$you got the $frac12$ part in both factors, which is wrong.






          share|cite|improve this answer









          $endgroup$



          It happens that$$frac{xlvert xrvert}2=xtimesfrac{lvert xrvert}2;$$you got the $frac12$ part in both factors, which is wrong.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 15:48









          José Carlos SantosJosé Carlos Santos

          156k22126227




          156k22126227












          • $begingroup$
            Add to that the missing $2$ in the denominator of the first term after applying product rule.
            $endgroup$
            – Shubham Johri
            Dec 2 '18 at 15:57


















          • $begingroup$
            Add to that the missing $2$ in the denominator of the first term after applying product rule.
            $endgroup$
            – Shubham Johri
            Dec 2 '18 at 15:57
















          $begingroup$
          Add to that the missing $2$ in the denominator of the first term after applying product rule.
          $endgroup$
          – Shubham Johri
          Dec 2 '18 at 15:57




          $begingroup$
          Add to that the missing $2$ in the denominator of the first term after applying product rule.
          $endgroup$
          – Shubham Johri
          Dec 2 '18 at 15:57











          0












          $begingroup$

          Consider what the product of $x$ and $|x|$ should look like. You'd get a piecewise function which is continuous.
          It is $-x^2$ for $x<0$ and $x^2$ for $x>0$.



          Differentiate this along those intervals.
          You should get $2|x|$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hm I'll try this out, thanks!
            $endgroup$
            – James Ronald
            Dec 2 '18 at 16:40
















          0












          $begingroup$

          Consider what the product of $x$ and $|x|$ should look like. You'd get a piecewise function which is continuous.
          It is $-x^2$ for $x<0$ and $x^2$ for $x>0$.



          Differentiate this along those intervals.
          You should get $2|x|$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hm I'll try this out, thanks!
            $endgroup$
            – James Ronald
            Dec 2 '18 at 16:40














          0












          0








          0





          $begingroup$

          Consider what the product of $x$ and $|x|$ should look like. You'd get a piecewise function which is continuous.
          It is $-x^2$ for $x<0$ and $x^2$ for $x>0$.



          Differentiate this along those intervals.
          You should get $2|x|$.






          share|cite|improve this answer









          $endgroup$



          Consider what the product of $x$ and $|x|$ should look like. You'd get a piecewise function which is continuous.
          It is $-x^2$ for $x<0$ and $x^2$ for $x>0$.



          Differentiate this along those intervals.
          You should get $2|x|$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 15:57









          Ryan GouldenRyan Goulden

          371110




          371110












          • $begingroup$
            Hm I'll try this out, thanks!
            $endgroup$
            – James Ronald
            Dec 2 '18 at 16:40


















          • $begingroup$
            Hm I'll try this out, thanks!
            $endgroup$
            – James Ronald
            Dec 2 '18 at 16:40
















          $begingroup$
          Hm I'll try this out, thanks!
          $endgroup$
          – James Ronald
          Dec 2 '18 at 16:40




          $begingroup$
          Hm I'll try this out, thanks!
          $endgroup$
          – James Ronald
          Dec 2 '18 at 16:40











          0












          $begingroup$

          False statement in your argument:
          $$
          frac{x|x|}{2}=frac{x}{2}frac{|x|}{2}
          $$



          Get rid of the factor $1/2$, but note that you cannot apply the product rule at $0$, because $|x|$ is not differentiable there.



          With the product rule you have, for $xne0$,
          $$
          F'(x)=frac{1}{2}(x|x|)'=frac{1}{2}left(|x|+xfrac{|x|}{x}right)=frac{1}{2}2|x|=|x|
          $$



          Now we also have to compute $F'(0)$:
          $$
          F'(0)=lim_{xto0}frac{F(x)-F(0)}{x}=lim_{xto0}frac{x|x|/2}{x}=
          lim_{xto0}frac{|x|}{2}=0
          $$
          Hence $F'(x)=|x|$ for every $x$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks so much for showing the rest of the steps as well, I understand now. Thank you!
            $endgroup$
            – James Ronald
            Dec 2 '18 at 16:40
















          0












          $begingroup$

          False statement in your argument:
          $$
          frac{x|x|}{2}=frac{x}{2}frac{|x|}{2}
          $$



          Get rid of the factor $1/2$, but note that you cannot apply the product rule at $0$, because $|x|$ is not differentiable there.



          With the product rule you have, for $xne0$,
          $$
          F'(x)=frac{1}{2}(x|x|)'=frac{1}{2}left(|x|+xfrac{|x|}{x}right)=frac{1}{2}2|x|=|x|
          $$



          Now we also have to compute $F'(0)$:
          $$
          F'(0)=lim_{xto0}frac{F(x)-F(0)}{x}=lim_{xto0}frac{x|x|/2}{x}=
          lim_{xto0}frac{|x|}{2}=0
          $$
          Hence $F'(x)=|x|$ for every $x$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks so much for showing the rest of the steps as well, I understand now. Thank you!
            $endgroup$
            – James Ronald
            Dec 2 '18 at 16:40














          0












          0








          0





          $begingroup$

          False statement in your argument:
          $$
          frac{x|x|}{2}=frac{x}{2}frac{|x|}{2}
          $$



          Get rid of the factor $1/2$, but note that you cannot apply the product rule at $0$, because $|x|$ is not differentiable there.



          With the product rule you have, for $xne0$,
          $$
          F'(x)=frac{1}{2}(x|x|)'=frac{1}{2}left(|x|+xfrac{|x|}{x}right)=frac{1}{2}2|x|=|x|
          $$



          Now we also have to compute $F'(0)$:
          $$
          F'(0)=lim_{xto0}frac{F(x)-F(0)}{x}=lim_{xto0}frac{x|x|/2}{x}=
          lim_{xto0}frac{|x|}{2}=0
          $$
          Hence $F'(x)=|x|$ for every $x$.






          share|cite|improve this answer









          $endgroup$



          False statement in your argument:
          $$
          frac{x|x|}{2}=frac{x}{2}frac{|x|}{2}
          $$



          Get rid of the factor $1/2$, but note that you cannot apply the product rule at $0$, because $|x|$ is not differentiable there.



          With the product rule you have, for $xne0$,
          $$
          F'(x)=frac{1}{2}(x|x|)'=frac{1}{2}left(|x|+xfrac{|x|}{x}right)=frac{1}{2}2|x|=|x|
          $$



          Now we also have to compute $F'(0)$:
          $$
          F'(0)=lim_{xto0}frac{F(x)-F(0)}{x}=lim_{xto0}frac{x|x|/2}{x}=
          lim_{xto0}frac{|x|}{2}=0
          $$
          Hence $F'(x)=|x|$ for every $x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 16:08









          egregegreg

          180k1485202




          180k1485202












          • $begingroup$
            Thanks so much for showing the rest of the steps as well, I understand now. Thank you!
            $endgroup$
            – James Ronald
            Dec 2 '18 at 16:40


















          • $begingroup$
            Thanks so much for showing the rest of the steps as well, I understand now. Thank you!
            $endgroup$
            – James Ronald
            Dec 2 '18 at 16:40
















          $begingroup$
          Thanks so much for showing the rest of the steps as well, I understand now. Thank you!
          $endgroup$
          – James Ronald
          Dec 2 '18 at 16:40




          $begingroup$
          Thanks so much for showing the rest of the steps as well, I understand now. Thank you!
          $endgroup$
          – James Ronald
          Dec 2 '18 at 16:40


















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