Confusion on multivariable partial differentiation from textbook












1












$begingroup$


I am very confused about a use of the partial differential multivariable chain rule. The textbook first describes the formula:



$frac{partial f}{partial s} = frac{partial f}{partial x}frac{partial x}{partial s} + frac{partial f}{partial y}frac{partial y}{partial s} + frac{partial f}{partial z}frac{partial z}{partial s}$



But then, as shown in the image provided, they set $s = x$ and call x to be a function of x, y to be a function of y, z to be a function of x and y, and make:



$f = x^2 + y^2 + z^2 = 4$.



But then on one side they claim:



$ frac{partial f}{partial x} = 0$ *



And on the other side they have:



$frac{partial f}{partial x}frac{partial x}{partial x} + frac{partial f}{partial y}frac{partial y}{partial x} + frac{partial f}{partial x}frac{partial z}{partial x} = (2x)(0) + (frac{partial f}{partial y})(0) + (2z)(frac{partial z}{partial x})$ (1)



So then $ frac{partial f}{partial x} = 2x$, but this is in direct contradiction with (*) above. This leads to large issues in equation (1), since the $ frac{partial f}{partial x}$ should cancel from both sides, but it does not for some reason.



Can someone please explain what is going on?



textbook image










share|cite|improve this question









$endgroup$












  • $begingroup$
    They're doing this implicitly.
    $endgroup$
    – Randall
    Dec 16 '18 at 2:20










  • $begingroup$
    Note that the book says $z=z(x,y)$.
    $endgroup$
    – manooooh
    Dec 16 '18 at 2:32










  • $begingroup$
    What you have written in your question is not what it says in the textbook. Perhaps if you compare them carefully, things will be plainer to you.
    $endgroup$
    – saulspatz
    Dec 16 '18 at 3:20










  • $begingroup$
    @saulspatz Thanks, I will try to understand the issues in my interpretation. However, my main question is why ∂f/∂x achieves different values on LHS and RHS.
    $endgroup$
    – A. Remorov
    Dec 16 '18 at 3:47










  • $begingroup$
    It doesn't. There are two different methods of computing it, and of course, they have to be equal.
    $endgroup$
    – saulspatz
    Dec 16 '18 at 3:51
















1












$begingroup$


I am very confused about a use of the partial differential multivariable chain rule. The textbook first describes the formula:



$frac{partial f}{partial s} = frac{partial f}{partial x}frac{partial x}{partial s} + frac{partial f}{partial y}frac{partial y}{partial s} + frac{partial f}{partial z}frac{partial z}{partial s}$



But then, as shown in the image provided, they set $s = x$ and call x to be a function of x, y to be a function of y, z to be a function of x and y, and make:



$f = x^2 + y^2 + z^2 = 4$.



But then on one side they claim:



$ frac{partial f}{partial x} = 0$ *



And on the other side they have:



$frac{partial f}{partial x}frac{partial x}{partial x} + frac{partial f}{partial y}frac{partial y}{partial x} + frac{partial f}{partial x}frac{partial z}{partial x} = (2x)(0) + (frac{partial f}{partial y})(0) + (2z)(frac{partial z}{partial x})$ (1)



So then $ frac{partial f}{partial x} = 2x$, but this is in direct contradiction with (*) above. This leads to large issues in equation (1), since the $ frac{partial f}{partial x}$ should cancel from both sides, but it does not for some reason.



Can someone please explain what is going on?



textbook image










share|cite|improve this question









$endgroup$












  • $begingroup$
    They're doing this implicitly.
    $endgroup$
    – Randall
    Dec 16 '18 at 2:20










  • $begingroup$
    Note that the book says $z=z(x,y)$.
    $endgroup$
    – manooooh
    Dec 16 '18 at 2:32










  • $begingroup$
    What you have written in your question is not what it says in the textbook. Perhaps if you compare them carefully, things will be plainer to you.
    $endgroup$
    – saulspatz
    Dec 16 '18 at 3:20










  • $begingroup$
    @saulspatz Thanks, I will try to understand the issues in my interpretation. However, my main question is why ∂f/∂x achieves different values on LHS and RHS.
    $endgroup$
    – A. Remorov
    Dec 16 '18 at 3:47










  • $begingroup$
    It doesn't. There are two different methods of computing it, and of course, they have to be equal.
    $endgroup$
    – saulspatz
    Dec 16 '18 at 3:51














1












1








1





$begingroup$


I am very confused about a use of the partial differential multivariable chain rule. The textbook first describes the formula:



$frac{partial f}{partial s} = frac{partial f}{partial x}frac{partial x}{partial s} + frac{partial f}{partial y}frac{partial y}{partial s} + frac{partial f}{partial z}frac{partial z}{partial s}$



But then, as shown in the image provided, they set $s = x$ and call x to be a function of x, y to be a function of y, z to be a function of x and y, and make:



$f = x^2 + y^2 + z^2 = 4$.



But then on one side they claim:



$ frac{partial f}{partial x} = 0$ *



And on the other side they have:



$frac{partial f}{partial x}frac{partial x}{partial x} + frac{partial f}{partial y}frac{partial y}{partial x} + frac{partial f}{partial x}frac{partial z}{partial x} = (2x)(0) + (frac{partial f}{partial y})(0) + (2z)(frac{partial z}{partial x})$ (1)



So then $ frac{partial f}{partial x} = 2x$, but this is in direct contradiction with (*) above. This leads to large issues in equation (1), since the $ frac{partial f}{partial x}$ should cancel from both sides, but it does not for some reason.



Can someone please explain what is going on?



textbook image










share|cite|improve this question









$endgroup$




I am very confused about a use of the partial differential multivariable chain rule. The textbook first describes the formula:



$frac{partial f}{partial s} = frac{partial f}{partial x}frac{partial x}{partial s} + frac{partial f}{partial y}frac{partial y}{partial s} + frac{partial f}{partial z}frac{partial z}{partial s}$



But then, as shown in the image provided, they set $s = x$ and call x to be a function of x, y to be a function of y, z to be a function of x and y, and make:



$f = x^2 + y^2 + z^2 = 4$.



But then on one side they claim:



$ frac{partial f}{partial x} = 0$ *



And on the other side they have:



$frac{partial f}{partial x}frac{partial x}{partial x} + frac{partial f}{partial y}frac{partial y}{partial x} + frac{partial f}{partial x}frac{partial z}{partial x} = (2x)(0) + (frac{partial f}{partial y})(0) + (2z)(frac{partial z}{partial x})$ (1)



So then $ frac{partial f}{partial x} = 2x$, but this is in direct contradiction with (*) above. This leads to large issues in equation (1), since the $ frac{partial f}{partial x}$ should cancel from both sides, but it does not for some reason.



Can someone please explain what is going on?



textbook image







multivariable-calculus partial-derivative chain-rule






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share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 2:09









A. RemorovA. Remorov

82




82












  • $begingroup$
    They're doing this implicitly.
    $endgroup$
    – Randall
    Dec 16 '18 at 2:20










  • $begingroup$
    Note that the book says $z=z(x,y)$.
    $endgroup$
    – manooooh
    Dec 16 '18 at 2:32










  • $begingroup$
    What you have written in your question is not what it says in the textbook. Perhaps if you compare them carefully, things will be plainer to you.
    $endgroup$
    – saulspatz
    Dec 16 '18 at 3:20










  • $begingroup$
    @saulspatz Thanks, I will try to understand the issues in my interpretation. However, my main question is why ∂f/∂x achieves different values on LHS and RHS.
    $endgroup$
    – A. Remorov
    Dec 16 '18 at 3:47










  • $begingroup$
    It doesn't. There are two different methods of computing it, and of course, they have to be equal.
    $endgroup$
    – saulspatz
    Dec 16 '18 at 3:51


















  • $begingroup$
    They're doing this implicitly.
    $endgroup$
    – Randall
    Dec 16 '18 at 2:20










  • $begingroup$
    Note that the book says $z=z(x,y)$.
    $endgroup$
    – manooooh
    Dec 16 '18 at 2:32










  • $begingroup$
    What you have written in your question is not what it says in the textbook. Perhaps if you compare them carefully, things will be plainer to you.
    $endgroup$
    – saulspatz
    Dec 16 '18 at 3:20










  • $begingroup$
    @saulspatz Thanks, I will try to understand the issues in my interpretation. However, my main question is why ∂f/∂x achieves different values on LHS and RHS.
    $endgroup$
    – A. Remorov
    Dec 16 '18 at 3:47










  • $begingroup$
    It doesn't. There are two different methods of computing it, and of course, they have to be equal.
    $endgroup$
    – saulspatz
    Dec 16 '18 at 3:51
















$begingroup$
They're doing this implicitly.
$endgroup$
– Randall
Dec 16 '18 at 2:20




$begingroup$
They're doing this implicitly.
$endgroup$
– Randall
Dec 16 '18 at 2:20












$begingroup$
Note that the book says $z=z(x,y)$.
$endgroup$
– manooooh
Dec 16 '18 at 2:32




$begingroup$
Note that the book says $z=z(x,y)$.
$endgroup$
– manooooh
Dec 16 '18 at 2:32












$begingroup$
What you have written in your question is not what it says in the textbook. Perhaps if you compare them carefully, things will be plainer to you.
$endgroup$
– saulspatz
Dec 16 '18 at 3:20




$begingroup$
What you have written in your question is not what it says in the textbook. Perhaps if you compare them carefully, things will be plainer to you.
$endgroup$
– saulspatz
Dec 16 '18 at 3:20












$begingroup$
@saulspatz Thanks, I will try to understand the issues in my interpretation. However, my main question is why ∂f/∂x achieves different values on LHS and RHS.
$endgroup$
– A. Remorov
Dec 16 '18 at 3:47




$begingroup$
@saulspatz Thanks, I will try to understand the issues in my interpretation. However, my main question is why ∂f/∂x achieves different values on LHS and RHS.
$endgroup$
– A. Remorov
Dec 16 '18 at 3:47












$begingroup$
It doesn't. There are two different methods of computing it, and of course, they have to be equal.
$endgroup$
– saulspatz
Dec 16 '18 at 3:51




$begingroup$
It doesn't. There are two different methods of computing it, and of course, they have to be equal.
$endgroup$
– saulspatz
Dec 16 '18 at 3:51










2 Answers
2






active

oldest

votes


















0












$begingroup$

In my opinion, it's very bad notation to use $partial f/partial x$ and $f_x$ for two different things, as they do. And in general it often leads to confusion when one uses the same name for the quantity that depends on some other quantities as for the function which describes this dependence.



If you want to clean it up, write $z=Z(y,z)$ instead of $z=z(x,y)$ (so that the function's name is $Z$, different from the variable name $z$), and write
$$
F(x,y) = f(x,y,Z(x,y)) = x^2 + y^2 + Z(x,y)^2
$$

for the composite function, which is actually the constant function $F(x,y)=4$ since that's how $Z(x,y)$ is defined. This function $F$ is a function of two variables, so it's clearly not the same as the three-variable function $f$, so it should have a different name.



Then what they denote by $partial f/partial x$ is really $partial F/partial x$, and it's zero on the one hand (since $F$ is constant), and given by the chain rule on the other hand (since $F$ is a composite function):
$$
begin{split}
0 = F_x(x,y)
& = f_x(x,y,Z(x,y)) + f_z(x,y,Z(x,y)) , Z_x(x,y)
\
& = 2x + 2 Z(x,y) , Z_x(x,y)
.
end{split}
$$

This shows that
$$
Z_x(x,y)
= - frac{f_x(x,y,Z(x,y))}{f_z(x,y,Z(x,y))}
= - frac{x}{Z(x,y)}
.
$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    We have $$x^2+y^2+z^2=4.tag{1}$$ As the book says, we can view this equation as defining $z$ as a function of the independent variables $x$ and $y$. Obviously, there is more than one possible function $z.$ Now if we consider the explicit function $f(x,y,z) = x^2+y^2+z^2$ where $z$ is the implicit function we just discussed, then we have$${partial fover partial x} = 2xcdot1+2ycdot0+2zcdot{partial zoverpartial x}tag{2}$$ This is true regardless of $(1)$.



    As I understand your question, you are asking, "If we know from $(1)$ that $f_x=0$ why don't we have $0$ instead of $2x$ on the right-hand side of $(2)?$" The short answer is that it would be premature. Equation $(2)$ expresses a general rule, which is true whenever $z$ is a function of $x$ and $y$. It has nothing to do with an implicitly defined function. You could substitute $z=sin(xy)$ and $(2)$ would be true.



    Now, we know from $(1)$ that with this particular value of $z,$ we have $f_x=0,$ so we can substitute this on the left-hand side of $(2)$ and solve for ${partial zoverpartial x}.$



    I'm sure there's a better way to express what I'm trying to say, but I hope this will make things a bit clearer to you.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I'm not sure this addresses the fundamental confusion, which I think is the following: if $f(x,y,z)=x^2+y^2+z^2$, then obviously $partial f/partial x=2x$, so how come $partial f/partial x=0$ all of a sudden? As I've tried to explain in my answer, this is because the same letter $f$ (confusingly) is used to denote two different functions.
      $endgroup$
      – Hans Lundmark
      Dec 16 '18 at 10:42










    • $begingroup$
      It may be obvious, but it isn't true. If $z$ is a function of $x$ and $y$ then equation $(2)$ holds. If $z$ happens to be such that equation $(1)$ holds, then $f_x=0.$ Your formula is correct if $x,y,$ and $z$ are independent variables, but they aren't.
      $endgroup$
      – saulspatz
      Dec 16 '18 at 14:40










    • $begingroup$
      My point is that the source of the confusion is this kind of notation, where you have to read from the context what depends on what. In the equation $f(x,y,z)=x^2+y^2+z^2=4$ one is defining a three-variable function $f(x,y,z)$, and looking at a level set of that function. But the symbol $f$ still denotes a three-variable function, whose partial derivative $partial f/partial x$ or $f_x$ or however you denote it is indisputably equal to $2x$. In the image in the question, they have indeed written $f_x$ in one place and in the next line written that it equals $2x$. (cont.)
      $endgroup$
      – Hans Lundmark
      Dec 16 '18 at 14:51












    • $begingroup$
      But in the same place they start out $0 = partial f/partial x = dots$, where the same symbol $f$ now stands for another function. And that function's partial with respect to $x$ is zero. So it's no wonder (to me) that students get confused.
      $endgroup$
      – Hans Lundmark
      Dec 16 '18 at 14:53










    • $begingroup$
      @HansLundmark Yes, I agree that it would be better if they used different letters, but this kind of thing is ubiquitous in calculus. In the chain rule $${dyover dx}={dyover du}{duover dx},$$ $y$ means different things on the two sides of the equation.
      $endgroup$
      – saulspatz
      Dec 16 '18 at 15:15











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    2 Answers
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    2 Answers
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    active

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    active

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    active

    oldest

    votes









    0












    $begingroup$

    In my opinion, it's very bad notation to use $partial f/partial x$ and $f_x$ for two different things, as they do. And in general it often leads to confusion when one uses the same name for the quantity that depends on some other quantities as for the function which describes this dependence.



    If you want to clean it up, write $z=Z(y,z)$ instead of $z=z(x,y)$ (so that the function's name is $Z$, different from the variable name $z$), and write
    $$
    F(x,y) = f(x,y,Z(x,y)) = x^2 + y^2 + Z(x,y)^2
    $$

    for the composite function, which is actually the constant function $F(x,y)=4$ since that's how $Z(x,y)$ is defined. This function $F$ is a function of two variables, so it's clearly not the same as the three-variable function $f$, so it should have a different name.



    Then what they denote by $partial f/partial x$ is really $partial F/partial x$, and it's zero on the one hand (since $F$ is constant), and given by the chain rule on the other hand (since $F$ is a composite function):
    $$
    begin{split}
    0 = F_x(x,y)
    & = f_x(x,y,Z(x,y)) + f_z(x,y,Z(x,y)) , Z_x(x,y)
    \
    & = 2x + 2 Z(x,y) , Z_x(x,y)
    .
    end{split}
    $$

    This shows that
    $$
    Z_x(x,y)
    = - frac{f_x(x,y,Z(x,y))}{f_z(x,y,Z(x,y))}
    = - frac{x}{Z(x,y)}
    .
    $$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      In my opinion, it's very bad notation to use $partial f/partial x$ and $f_x$ for two different things, as they do. And in general it often leads to confusion when one uses the same name for the quantity that depends on some other quantities as for the function which describes this dependence.



      If you want to clean it up, write $z=Z(y,z)$ instead of $z=z(x,y)$ (so that the function's name is $Z$, different from the variable name $z$), and write
      $$
      F(x,y) = f(x,y,Z(x,y)) = x^2 + y^2 + Z(x,y)^2
      $$

      for the composite function, which is actually the constant function $F(x,y)=4$ since that's how $Z(x,y)$ is defined. This function $F$ is a function of two variables, so it's clearly not the same as the three-variable function $f$, so it should have a different name.



      Then what they denote by $partial f/partial x$ is really $partial F/partial x$, and it's zero on the one hand (since $F$ is constant), and given by the chain rule on the other hand (since $F$ is a composite function):
      $$
      begin{split}
      0 = F_x(x,y)
      & = f_x(x,y,Z(x,y)) + f_z(x,y,Z(x,y)) , Z_x(x,y)
      \
      & = 2x + 2 Z(x,y) , Z_x(x,y)
      .
      end{split}
      $$

      This shows that
      $$
      Z_x(x,y)
      = - frac{f_x(x,y,Z(x,y))}{f_z(x,y,Z(x,y))}
      = - frac{x}{Z(x,y)}
      .
      $$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        In my opinion, it's very bad notation to use $partial f/partial x$ and $f_x$ for two different things, as they do. And in general it often leads to confusion when one uses the same name for the quantity that depends on some other quantities as for the function which describes this dependence.



        If you want to clean it up, write $z=Z(y,z)$ instead of $z=z(x,y)$ (so that the function's name is $Z$, different from the variable name $z$), and write
        $$
        F(x,y) = f(x,y,Z(x,y)) = x^2 + y^2 + Z(x,y)^2
        $$

        for the composite function, which is actually the constant function $F(x,y)=4$ since that's how $Z(x,y)$ is defined. This function $F$ is a function of two variables, so it's clearly not the same as the three-variable function $f$, so it should have a different name.



        Then what they denote by $partial f/partial x$ is really $partial F/partial x$, and it's zero on the one hand (since $F$ is constant), and given by the chain rule on the other hand (since $F$ is a composite function):
        $$
        begin{split}
        0 = F_x(x,y)
        & = f_x(x,y,Z(x,y)) + f_z(x,y,Z(x,y)) , Z_x(x,y)
        \
        & = 2x + 2 Z(x,y) , Z_x(x,y)
        .
        end{split}
        $$

        This shows that
        $$
        Z_x(x,y)
        = - frac{f_x(x,y,Z(x,y))}{f_z(x,y,Z(x,y))}
        = - frac{x}{Z(x,y)}
        .
        $$






        share|cite|improve this answer











        $endgroup$



        In my opinion, it's very bad notation to use $partial f/partial x$ and $f_x$ for two different things, as they do. And in general it often leads to confusion when one uses the same name for the quantity that depends on some other quantities as for the function which describes this dependence.



        If you want to clean it up, write $z=Z(y,z)$ instead of $z=z(x,y)$ (so that the function's name is $Z$, different from the variable name $z$), and write
        $$
        F(x,y) = f(x,y,Z(x,y)) = x^2 + y^2 + Z(x,y)^2
        $$

        for the composite function, which is actually the constant function $F(x,y)=4$ since that's how $Z(x,y)$ is defined. This function $F$ is a function of two variables, so it's clearly not the same as the three-variable function $f$, so it should have a different name.



        Then what they denote by $partial f/partial x$ is really $partial F/partial x$, and it's zero on the one hand (since $F$ is constant), and given by the chain rule on the other hand (since $F$ is a composite function):
        $$
        begin{split}
        0 = F_x(x,y)
        & = f_x(x,y,Z(x,y)) + f_z(x,y,Z(x,y)) , Z_x(x,y)
        \
        & = 2x + 2 Z(x,y) , Z_x(x,y)
        .
        end{split}
        $$

        This shows that
        $$
        Z_x(x,y)
        = - frac{f_x(x,y,Z(x,y))}{f_z(x,y,Z(x,y))}
        = - frac{x}{Z(x,y)}
        .
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 14:36

























        answered Dec 16 '18 at 10:38









        Hans LundmarkHans Lundmark

        35.6k564115




        35.6k564115























            0












            $begingroup$

            We have $$x^2+y^2+z^2=4.tag{1}$$ As the book says, we can view this equation as defining $z$ as a function of the independent variables $x$ and $y$. Obviously, there is more than one possible function $z.$ Now if we consider the explicit function $f(x,y,z) = x^2+y^2+z^2$ where $z$ is the implicit function we just discussed, then we have$${partial fover partial x} = 2xcdot1+2ycdot0+2zcdot{partial zoverpartial x}tag{2}$$ This is true regardless of $(1)$.



            As I understand your question, you are asking, "If we know from $(1)$ that $f_x=0$ why don't we have $0$ instead of $2x$ on the right-hand side of $(2)?$" The short answer is that it would be premature. Equation $(2)$ expresses a general rule, which is true whenever $z$ is a function of $x$ and $y$. It has nothing to do with an implicitly defined function. You could substitute $z=sin(xy)$ and $(2)$ would be true.



            Now, we know from $(1)$ that with this particular value of $z,$ we have $f_x=0,$ so we can substitute this on the left-hand side of $(2)$ and solve for ${partial zoverpartial x}.$



            I'm sure there's a better way to express what I'm trying to say, but I hope this will make things a bit clearer to you.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I'm not sure this addresses the fundamental confusion, which I think is the following: if $f(x,y,z)=x^2+y^2+z^2$, then obviously $partial f/partial x=2x$, so how come $partial f/partial x=0$ all of a sudden? As I've tried to explain in my answer, this is because the same letter $f$ (confusingly) is used to denote two different functions.
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 10:42










            • $begingroup$
              It may be obvious, but it isn't true. If $z$ is a function of $x$ and $y$ then equation $(2)$ holds. If $z$ happens to be such that equation $(1)$ holds, then $f_x=0.$ Your formula is correct if $x,y,$ and $z$ are independent variables, but they aren't.
              $endgroup$
              – saulspatz
              Dec 16 '18 at 14:40










            • $begingroup$
              My point is that the source of the confusion is this kind of notation, where you have to read from the context what depends on what. In the equation $f(x,y,z)=x^2+y^2+z^2=4$ one is defining a three-variable function $f(x,y,z)$, and looking at a level set of that function. But the symbol $f$ still denotes a three-variable function, whose partial derivative $partial f/partial x$ or $f_x$ or however you denote it is indisputably equal to $2x$. In the image in the question, they have indeed written $f_x$ in one place and in the next line written that it equals $2x$. (cont.)
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 14:51












            • $begingroup$
              But in the same place they start out $0 = partial f/partial x = dots$, where the same symbol $f$ now stands for another function. And that function's partial with respect to $x$ is zero. So it's no wonder (to me) that students get confused.
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 14:53










            • $begingroup$
              @HansLundmark Yes, I agree that it would be better if they used different letters, but this kind of thing is ubiquitous in calculus. In the chain rule $${dyover dx}={dyover du}{duover dx},$$ $y$ means different things on the two sides of the equation.
              $endgroup$
              – saulspatz
              Dec 16 '18 at 15:15
















            0












            $begingroup$

            We have $$x^2+y^2+z^2=4.tag{1}$$ As the book says, we can view this equation as defining $z$ as a function of the independent variables $x$ and $y$. Obviously, there is more than one possible function $z.$ Now if we consider the explicit function $f(x,y,z) = x^2+y^2+z^2$ where $z$ is the implicit function we just discussed, then we have$${partial fover partial x} = 2xcdot1+2ycdot0+2zcdot{partial zoverpartial x}tag{2}$$ This is true regardless of $(1)$.



            As I understand your question, you are asking, "If we know from $(1)$ that $f_x=0$ why don't we have $0$ instead of $2x$ on the right-hand side of $(2)?$" The short answer is that it would be premature. Equation $(2)$ expresses a general rule, which is true whenever $z$ is a function of $x$ and $y$. It has nothing to do with an implicitly defined function. You could substitute $z=sin(xy)$ and $(2)$ would be true.



            Now, we know from $(1)$ that with this particular value of $z,$ we have $f_x=0,$ so we can substitute this on the left-hand side of $(2)$ and solve for ${partial zoverpartial x}.$



            I'm sure there's a better way to express what I'm trying to say, but I hope this will make things a bit clearer to you.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I'm not sure this addresses the fundamental confusion, which I think is the following: if $f(x,y,z)=x^2+y^2+z^2$, then obviously $partial f/partial x=2x$, so how come $partial f/partial x=0$ all of a sudden? As I've tried to explain in my answer, this is because the same letter $f$ (confusingly) is used to denote two different functions.
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 10:42










            • $begingroup$
              It may be obvious, but it isn't true. If $z$ is a function of $x$ and $y$ then equation $(2)$ holds. If $z$ happens to be such that equation $(1)$ holds, then $f_x=0.$ Your formula is correct if $x,y,$ and $z$ are independent variables, but they aren't.
              $endgroup$
              – saulspatz
              Dec 16 '18 at 14:40










            • $begingroup$
              My point is that the source of the confusion is this kind of notation, where you have to read from the context what depends on what. In the equation $f(x,y,z)=x^2+y^2+z^2=4$ one is defining a three-variable function $f(x,y,z)$, and looking at a level set of that function. But the symbol $f$ still denotes a three-variable function, whose partial derivative $partial f/partial x$ or $f_x$ or however you denote it is indisputably equal to $2x$. In the image in the question, they have indeed written $f_x$ in one place and in the next line written that it equals $2x$. (cont.)
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 14:51












            • $begingroup$
              But in the same place they start out $0 = partial f/partial x = dots$, where the same symbol $f$ now stands for another function. And that function's partial with respect to $x$ is zero. So it's no wonder (to me) that students get confused.
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 14:53










            • $begingroup$
              @HansLundmark Yes, I agree that it would be better if they used different letters, but this kind of thing is ubiquitous in calculus. In the chain rule $${dyover dx}={dyover du}{duover dx},$$ $y$ means different things on the two sides of the equation.
              $endgroup$
              – saulspatz
              Dec 16 '18 at 15:15














            0












            0








            0





            $begingroup$

            We have $$x^2+y^2+z^2=4.tag{1}$$ As the book says, we can view this equation as defining $z$ as a function of the independent variables $x$ and $y$. Obviously, there is more than one possible function $z.$ Now if we consider the explicit function $f(x,y,z) = x^2+y^2+z^2$ where $z$ is the implicit function we just discussed, then we have$${partial fover partial x} = 2xcdot1+2ycdot0+2zcdot{partial zoverpartial x}tag{2}$$ This is true regardless of $(1)$.



            As I understand your question, you are asking, "If we know from $(1)$ that $f_x=0$ why don't we have $0$ instead of $2x$ on the right-hand side of $(2)?$" The short answer is that it would be premature. Equation $(2)$ expresses a general rule, which is true whenever $z$ is a function of $x$ and $y$. It has nothing to do with an implicitly defined function. You could substitute $z=sin(xy)$ and $(2)$ would be true.



            Now, we know from $(1)$ that with this particular value of $z,$ we have $f_x=0,$ so we can substitute this on the left-hand side of $(2)$ and solve for ${partial zoverpartial x}.$



            I'm sure there's a better way to express what I'm trying to say, but I hope this will make things a bit clearer to you.






            share|cite|improve this answer









            $endgroup$



            We have $$x^2+y^2+z^2=4.tag{1}$$ As the book says, we can view this equation as defining $z$ as a function of the independent variables $x$ and $y$. Obviously, there is more than one possible function $z.$ Now if we consider the explicit function $f(x,y,z) = x^2+y^2+z^2$ where $z$ is the implicit function we just discussed, then we have$${partial fover partial x} = 2xcdot1+2ycdot0+2zcdot{partial zoverpartial x}tag{2}$$ This is true regardless of $(1)$.



            As I understand your question, you are asking, "If we know from $(1)$ that $f_x=0$ why don't we have $0$ instead of $2x$ on the right-hand side of $(2)?$" The short answer is that it would be premature. Equation $(2)$ expresses a general rule, which is true whenever $z$ is a function of $x$ and $y$. It has nothing to do with an implicitly defined function. You could substitute $z=sin(xy)$ and $(2)$ would be true.



            Now, we know from $(1)$ that with this particular value of $z,$ we have $f_x=0,$ so we can substitute this on the left-hand side of $(2)$ and solve for ${partial zoverpartial x}.$



            I'm sure there's a better way to express what I'm trying to say, but I hope this will make things a bit clearer to you.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 16 '18 at 4:42









            saulspatzsaulspatz

            15.2k31331




            15.2k31331












            • $begingroup$
              I'm not sure this addresses the fundamental confusion, which I think is the following: if $f(x,y,z)=x^2+y^2+z^2$, then obviously $partial f/partial x=2x$, so how come $partial f/partial x=0$ all of a sudden? As I've tried to explain in my answer, this is because the same letter $f$ (confusingly) is used to denote two different functions.
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 10:42










            • $begingroup$
              It may be obvious, but it isn't true. If $z$ is a function of $x$ and $y$ then equation $(2)$ holds. If $z$ happens to be such that equation $(1)$ holds, then $f_x=0.$ Your formula is correct if $x,y,$ and $z$ are independent variables, but they aren't.
              $endgroup$
              – saulspatz
              Dec 16 '18 at 14:40










            • $begingroup$
              My point is that the source of the confusion is this kind of notation, where you have to read from the context what depends on what. In the equation $f(x,y,z)=x^2+y^2+z^2=4$ one is defining a three-variable function $f(x,y,z)$, and looking at a level set of that function. But the symbol $f$ still denotes a three-variable function, whose partial derivative $partial f/partial x$ or $f_x$ or however you denote it is indisputably equal to $2x$. In the image in the question, they have indeed written $f_x$ in one place and in the next line written that it equals $2x$. (cont.)
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 14:51












            • $begingroup$
              But in the same place they start out $0 = partial f/partial x = dots$, where the same symbol $f$ now stands for another function. And that function's partial with respect to $x$ is zero. So it's no wonder (to me) that students get confused.
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 14:53










            • $begingroup$
              @HansLundmark Yes, I agree that it would be better if they used different letters, but this kind of thing is ubiquitous in calculus. In the chain rule $${dyover dx}={dyover du}{duover dx},$$ $y$ means different things on the two sides of the equation.
              $endgroup$
              – saulspatz
              Dec 16 '18 at 15:15


















            • $begingroup$
              I'm not sure this addresses the fundamental confusion, which I think is the following: if $f(x,y,z)=x^2+y^2+z^2$, then obviously $partial f/partial x=2x$, so how come $partial f/partial x=0$ all of a sudden? As I've tried to explain in my answer, this is because the same letter $f$ (confusingly) is used to denote two different functions.
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 10:42










            • $begingroup$
              It may be obvious, but it isn't true. If $z$ is a function of $x$ and $y$ then equation $(2)$ holds. If $z$ happens to be such that equation $(1)$ holds, then $f_x=0.$ Your formula is correct if $x,y,$ and $z$ are independent variables, but they aren't.
              $endgroup$
              – saulspatz
              Dec 16 '18 at 14:40










            • $begingroup$
              My point is that the source of the confusion is this kind of notation, where you have to read from the context what depends on what. In the equation $f(x,y,z)=x^2+y^2+z^2=4$ one is defining a three-variable function $f(x,y,z)$, and looking at a level set of that function. But the symbol $f$ still denotes a three-variable function, whose partial derivative $partial f/partial x$ or $f_x$ or however you denote it is indisputably equal to $2x$. In the image in the question, they have indeed written $f_x$ in one place and in the next line written that it equals $2x$. (cont.)
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 14:51












            • $begingroup$
              But in the same place they start out $0 = partial f/partial x = dots$, where the same symbol $f$ now stands for another function. And that function's partial with respect to $x$ is zero. So it's no wonder (to me) that students get confused.
              $endgroup$
              – Hans Lundmark
              Dec 16 '18 at 14:53










            • $begingroup$
              @HansLundmark Yes, I agree that it would be better if they used different letters, but this kind of thing is ubiquitous in calculus. In the chain rule $${dyover dx}={dyover du}{duover dx},$$ $y$ means different things on the two sides of the equation.
              $endgroup$
              – saulspatz
              Dec 16 '18 at 15:15
















            $begingroup$
            I'm not sure this addresses the fundamental confusion, which I think is the following: if $f(x,y,z)=x^2+y^2+z^2$, then obviously $partial f/partial x=2x$, so how come $partial f/partial x=0$ all of a sudden? As I've tried to explain in my answer, this is because the same letter $f$ (confusingly) is used to denote two different functions.
            $endgroup$
            – Hans Lundmark
            Dec 16 '18 at 10:42




            $begingroup$
            I'm not sure this addresses the fundamental confusion, which I think is the following: if $f(x,y,z)=x^2+y^2+z^2$, then obviously $partial f/partial x=2x$, so how come $partial f/partial x=0$ all of a sudden? As I've tried to explain in my answer, this is because the same letter $f$ (confusingly) is used to denote two different functions.
            $endgroup$
            – Hans Lundmark
            Dec 16 '18 at 10:42












            $begingroup$
            It may be obvious, but it isn't true. If $z$ is a function of $x$ and $y$ then equation $(2)$ holds. If $z$ happens to be such that equation $(1)$ holds, then $f_x=0.$ Your formula is correct if $x,y,$ and $z$ are independent variables, but they aren't.
            $endgroup$
            – saulspatz
            Dec 16 '18 at 14:40




            $begingroup$
            It may be obvious, but it isn't true. If $z$ is a function of $x$ and $y$ then equation $(2)$ holds. If $z$ happens to be such that equation $(1)$ holds, then $f_x=0.$ Your formula is correct if $x,y,$ and $z$ are independent variables, but they aren't.
            $endgroup$
            – saulspatz
            Dec 16 '18 at 14:40












            $begingroup$
            My point is that the source of the confusion is this kind of notation, where you have to read from the context what depends on what. In the equation $f(x,y,z)=x^2+y^2+z^2=4$ one is defining a three-variable function $f(x,y,z)$, and looking at a level set of that function. But the symbol $f$ still denotes a three-variable function, whose partial derivative $partial f/partial x$ or $f_x$ or however you denote it is indisputably equal to $2x$. In the image in the question, they have indeed written $f_x$ in one place and in the next line written that it equals $2x$. (cont.)
            $endgroup$
            – Hans Lundmark
            Dec 16 '18 at 14:51






            $begingroup$
            My point is that the source of the confusion is this kind of notation, where you have to read from the context what depends on what. In the equation $f(x,y,z)=x^2+y^2+z^2=4$ one is defining a three-variable function $f(x,y,z)$, and looking at a level set of that function. But the symbol $f$ still denotes a three-variable function, whose partial derivative $partial f/partial x$ or $f_x$ or however you denote it is indisputably equal to $2x$. In the image in the question, they have indeed written $f_x$ in one place and in the next line written that it equals $2x$. (cont.)
            $endgroup$
            – Hans Lundmark
            Dec 16 '18 at 14:51














            $begingroup$
            But in the same place they start out $0 = partial f/partial x = dots$, where the same symbol $f$ now stands for another function. And that function's partial with respect to $x$ is zero. So it's no wonder (to me) that students get confused.
            $endgroup$
            – Hans Lundmark
            Dec 16 '18 at 14:53




            $begingroup$
            But in the same place they start out $0 = partial f/partial x = dots$, where the same symbol $f$ now stands for another function. And that function's partial with respect to $x$ is zero. So it's no wonder (to me) that students get confused.
            $endgroup$
            – Hans Lundmark
            Dec 16 '18 at 14:53












            $begingroup$
            @HansLundmark Yes, I agree that it would be better if they used different letters, but this kind of thing is ubiquitous in calculus. In the chain rule $${dyover dx}={dyover du}{duover dx},$$ $y$ means different things on the two sides of the equation.
            $endgroup$
            – saulspatz
            Dec 16 '18 at 15:15




            $begingroup$
            @HansLundmark Yes, I agree that it would be better if they used different letters, but this kind of thing is ubiquitous in calculus. In the chain rule $${dyover dx}={dyover du}{duover dx},$$ $y$ means different things on the two sides of the equation.
            $endgroup$
            – saulspatz
            Dec 16 '18 at 15:15


















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