Finding the error in an argument












4












$begingroup$


If $z=f(x,y)$ and $y=x^2$, then by the chain rule



$frac{partial z}{partial x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$



Therefore



$2xfrac{partial z}{partial y}=0$



and



$frac{partial z}{partial y}=0$



What is wrong with this argument?



I have a feeling that



1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and



2.) $frac{partial z}{partial y}$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.



How is my reasoning? I am pretty confused by this question.






calculus multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$












  • $begingroup$
    My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
    $endgroup$
    – BSplitter
    Apr 3 at 1:13








  • 1




    $begingroup$
    I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
    $endgroup$
    – BSplitter
    Apr 3 at 1:14










  • $begingroup$
    @BSplitter the problem itself poses this argument as incorrect, the object being to find out why
    $endgroup$
    – mathenthusiast
    Apr 3 at 1:17
















4












$begingroup$


If $z=f(x,y)$ and $y=x^2$, then by the chain rule



$frac{partial z}{partial x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$



Therefore



$2xfrac{partial z}{partial y}=0$



and



$frac{partial z}{partial y}=0$



What is wrong with this argument?



I have a feeling that



1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and



2.) $frac{partial z}{partial y}$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.



How is my reasoning? I am pretty confused by this question.






calculus multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$












  • $begingroup$
    My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
    $endgroup$
    – BSplitter
    Apr 3 at 1:13








  • 1




    $begingroup$
    I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
    $endgroup$
    – BSplitter
    Apr 3 at 1:14










  • $begingroup$
    @BSplitter the problem itself poses this argument as incorrect, the object being to find out why
    $endgroup$
    – mathenthusiast
    Apr 3 at 1:17














4












4








4





$begingroup$


If $z=f(x,y)$ and $y=x^2$, then by the chain rule



$frac{partial z}{partial x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$



Therefore



$2xfrac{partial z}{partial y}=0$



and



$frac{partial z}{partial y}=0$



What is wrong with this argument?



I have a feeling that



1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and



2.) $frac{partial z}{partial y}$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.



How is my reasoning? I am pretty confused by this question.






calculus multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$




If $z=f(x,y)$ and $y=x^2$, then by the chain rule



$frac{partial z}{partial x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$



Therefore



$2xfrac{partial z}{partial y}=0$



and



$frac{partial z}{partial y}=0$



What is wrong with this argument?



I have a feeling that



1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and



2.) $frac{partial z}{partial y}$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.



How is my reasoning? I am pretty confused by this question.






calculus multivariable-calculus partial-derivative




calculus multivariable-calculus partial-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 3 at 1:07







mathenthusiast

















asked Apr 3 at 0:52









mathenthusiastmathenthusiast

808




808












  • $begingroup$
    My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
    $endgroup$
    – BSplitter
    Apr 3 at 1:13








  • 1




    $begingroup$
    I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
    $endgroup$
    – BSplitter
    Apr 3 at 1:14










  • $begingroup$
    @BSplitter the problem itself poses this argument as incorrect, the object being to find out why
    $endgroup$
    – mathenthusiast
    Apr 3 at 1:17


















  • $begingroup$
    My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
    $endgroup$
    – BSplitter
    Apr 3 at 1:13








  • 1




    $begingroup$
    I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
    $endgroup$
    – BSplitter
    Apr 3 at 1:14










  • $begingroup$
    @BSplitter the problem itself poses this argument as incorrect, the object being to find out why
    $endgroup$
    – mathenthusiast
    Apr 3 at 1:17
















$begingroup$
My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
$endgroup$
– BSplitter
Apr 3 at 1:13






$begingroup$
My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
$endgroup$
– BSplitter
Apr 3 at 1:13






1




1




$begingroup$
I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
$endgroup$
– BSplitter
Apr 3 at 1:14




$begingroup$
I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
$endgroup$
– BSplitter
Apr 3 at 1:14












$begingroup$
@BSplitter the problem itself poses this argument as incorrect, the object being to find out why
$endgroup$
– mathenthusiast
Apr 3 at 1:17




$begingroup$
@BSplitter the problem itself poses this argument as incorrect, the object being to find out why
$endgroup$
– mathenthusiast
Apr 3 at 1:17










1 Answer
1






active

oldest

votes


















5












$begingroup$

Nothing wrong. Just change it into



$$frac{d z}{d x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$$



Note that that the first term is $frac{d z}{d x}$, which is different from $frac{partial z}{partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $frac{d z}{d x}$, which is NOT a partial derivative.



Actually, a better way to say this is that



$$left[frac{partial z}{partial x}right]_{y=x^2}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}.$$



Where I have clearly written down the restriction $y=x^2$.






share|cite|improve this answer









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    1 Answer
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    active

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    5












    $begingroup$

    Nothing wrong. Just change it into



    $$frac{d z}{d x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$$



    Note that that the first term is $frac{d z}{d x}$, which is different from $frac{partial z}{partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $frac{d z}{d x}$, which is NOT a partial derivative.



    Actually, a better way to say this is that



    $$left[frac{partial z}{partial x}right]_{y=x^2}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}.$$



    Where I have clearly written down the restriction $y=x^2$.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      Nothing wrong. Just change it into



      $$frac{d z}{d x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$$



      Note that that the first term is $frac{d z}{d x}$, which is different from $frac{partial z}{partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $frac{d z}{d x}$, which is NOT a partial derivative.



      Actually, a better way to say this is that



      $$left[frac{partial z}{partial x}right]_{y=x^2}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}.$$



      Where I have clearly written down the restriction $y=x^2$.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        Nothing wrong. Just change it into



        $$frac{d z}{d x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$$



        Note that that the first term is $frac{d z}{d x}$, which is different from $frac{partial z}{partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $frac{d z}{d x}$, which is NOT a partial derivative.



        Actually, a better way to say this is that



        $$left[frac{partial z}{partial x}right]_{y=x^2}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}.$$



        Where I have clearly written down the restriction $y=x^2$.






        share|cite|improve this answer









        $endgroup$



        Nothing wrong. Just change it into



        $$frac{d z}{d x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$$



        Note that that the first term is $frac{d z}{d x}$, which is different from $frac{partial z}{partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $frac{d z}{d x}$, which is NOT a partial derivative.



        Actually, a better way to say this is that



        $$left[frac{partial z}{partial x}right]_{y=x^2}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}.$$



        Where I have clearly written down the restriction $y=x^2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 3 at 1:21









        Holding ArthurHolding Arthur

        1,753417




        1,753417






























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