Partial Derivative with Respect to Multiple Variables












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If we take a multivariable function such as $w=f(x,y,z)=x^2+y^2+z^2$, I understand that we can take its partial derivative with respect to any one of its arguments, while the others stay unchanged. In this case, we can take the partial derivative with respect to $z$ as $frac{partial}{partial z}f(x,y,z)$. I also understand that we can take its total derivative which is with respect to all of its arguments, which can be expressed as $frac{dt}{dw}f(x,y,z)$.



My question: how do we represent the derivative with respect to some but not all of a multivariable function's variables? How is this derivative classified? I am tempted to think it is partial but I am not sure since all of the definitions I have seen give it with respect to a single variable. Is it a "partial" total derivative?










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  • 1




    $begingroup$
    Note: a total derivative would be something like $$frac{dw}{dt}=frac{partial w}{partial x}frac{dx}{dt}+frac{partial w}{partial y}frac{dy}{dt}+frac{partial w}{partial z}frac{dz}{dt}$$
    $endgroup$
    – John Doe
    Jan 4 at 21:58










  • $begingroup$
    See mixed partial derivatives
    $endgroup$
    – WaveX
    Jan 4 at 22:26










  • $begingroup$
    Your notation with the "total derivative" makes absolutely no sense. I assume you mean $frac d{dt} f(x(t),y(t),z(t))$ or, often, $frac d{dt} f(t,x(t),y(t),z(t))$.
    $endgroup$
    – Ted Shifrin
    Jan 4 at 23:12


















0












$begingroup$


If we take a multivariable function such as $w=f(x,y,z)=x^2+y^2+z^2$, I understand that we can take its partial derivative with respect to any one of its arguments, while the others stay unchanged. In this case, we can take the partial derivative with respect to $z$ as $frac{partial}{partial z}f(x,y,z)$. I also understand that we can take its total derivative which is with respect to all of its arguments, which can be expressed as $frac{dt}{dw}f(x,y,z)$.



My question: how do we represent the derivative with respect to some but not all of a multivariable function's variables? How is this derivative classified? I am tempted to think it is partial but I am not sure since all of the definitions I have seen give it with respect to a single variable. Is it a "partial" total derivative?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note: a total derivative would be something like $$frac{dw}{dt}=frac{partial w}{partial x}frac{dx}{dt}+frac{partial w}{partial y}frac{dy}{dt}+frac{partial w}{partial z}frac{dz}{dt}$$
    $endgroup$
    – John Doe
    Jan 4 at 21:58










  • $begingroup$
    See mixed partial derivatives
    $endgroup$
    – WaveX
    Jan 4 at 22:26










  • $begingroup$
    Your notation with the "total derivative" makes absolutely no sense. I assume you mean $frac d{dt} f(x(t),y(t),z(t))$ or, often, $frac d{dt} f(t,x(t),y(t),z(t))$.
    $endgroup$
    – Ted Shifrin
    Jan 4 at 23:12
















0












0








0





$begingroup$


If we take a multivariable function such as $w=f(x,y,z)=x^2+y^2+z^2$, I understand that we can take its partial derivative with respect to any one of its arguments, while the others stay unchanged. In this case, we can take the partial derivative with respect to $z$ as $frac{partial}{partial z}f(x,y,z)$. I also understand that we can take its total derivative which is with respect to all of its arguments, which can be expressed as $frac{dt}{dw}f(x,y,z)$.



My question: how do we represent the derivative with respect to some but not all of a multivariable function's variables? How is this derivative classified? I am tempted to think it is partial but I am not sure since all of the definitions I have seen give it with respect to a single variable. Is it a "partial" total derivative?










share|cite|improve this question











$endgroup$




If we take a multivariable function such as $w=f(x,y,z)=x^2+y^2+z^2$, I understand that we can take its partial derivative with respect to any one of its arguments, while the others stay unchanged. In this case, we can take the partial derivative with respect to $z$ as $frac{partial}{partial z}f(x,y,z)$. I also understand that we can take its total derivative which is with respect to all of its arguments, which can be expressed as $frac{dt}{dw}f(x,y,z)$.



My question: how do we represent the derivative with respect to some but not all of a multivariable function's variables? How is this derivative classified? I am tempted to think it is partial but I am not sure since all of the definitions I have seen give it with respect to a single variable. Is it a "partial" total derivative?







multivariable-calculus partial-derivative






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edited Jan 4 at 22:04







Gnumbertester

















asked Jan 4 at 21:51









GnumbertesterGnumbertester

668113




668113








  • 1




    $begingroup$
    Note: a total derivative would be something like $$frac{dw}{dt}=frac{partial w}{partial x}frac{dx}{dt}+frac{partial w}{partial y}frac{dy}{dt}+frac{partial w}{partial z}frac{dz}{dt}$$
    $endgroup$
    – John Doe
    Jan 4 at 21:58










  • $begingroup$
    See mixed partial derivatives
    $endgroup$
    – WaveX
    Jan 4 at 22:26










  • $begingroup$
    Your notation with the "total derivative" makes absolutely no sense. I assume you mean $frac d{dt} f(x(t),y(t),z(t))$ or, often, $frac d{dt} f(t,x(t),y(t),z(t))$.
    $endgroup$
    – Ted Shifrin
    Jan 4 at 23:12
















  • 1




    $begingroup$
    Note: a total derivative would be something like $$frac{dw}{dt}=frac{partial w}{partial x}frac{dx}{dt}+frac{partial w}{partial y}frac{dy}{dt}+frac{partial w}{partial z}frac{dz}{dt}$$
    $endgroup$
    – John Doe
    Jan 4 at 21:58










  • $begingroup$
    See mixed partial derivatives
    $endgroup$
    – WaveX
    Jan 4 at 22:26










  • $begingroup$
    Your notation with the "total derivative" makes absolutely no sense. I assume you mean $frac d{dt} f(x(t),y(t),z(t))$ or, often, $frac d{dt} f(t,x(t),y(t),z(t))$.
    $endgroup$
    – Ted Shifrin
    Jan 4 at 23:12










1




1




$begingroup$
Note: a total derivative would be something like $$frac{dw}{dt}=frac{partial w}{partial x}frac{dx}{dt}+frac{partial w}{partial y}frac{dy}{dt}+frac{partial w}{partial z}frac{dz}{dt}$$
$endgroup$
– John Doe
Jan 4 at 21:58




$begingroup$
Note: a total derivative would be something like $$frac{dw}{dt}=frac{partial w}{partial x}frac{dx}{dt}+frac{partial w}{partial y}frac{dy}{dt}+frac{partial w}{partial z}frac{dz}{dt}$$
$endgroup$
– John Doe
Jan 4 at 21:58












$begingroup$
See mixed partial derivatives
$endgroup$
– WaveX
Jan 4 at 22:26




$begingroup$
See mixed partial derivatives
$endgroup$
– WaveX
Jan 4 at 22:26












$begingroup$
Your notation with the "total derivative" makes absolutely no sense. I assume you mean $frac d{dt} f(x(t),y(t),z(t))$ or, often, $frac d{dt} f(t,x(t),y(t),z(t))$.
$endgroup$
– Ted Shifrin
Jan 4 at 23:12






$begingroup$
Your notation with the "total derivative" makes absolutely no sense. I assume you mean $frac d{dt} f(x(t),y(t),z(t))$ or, often, $frac d{dt} f(t,x(t),y(t),z(t))$.
$endgroup$
– Ted Shifrin
Jan 4 at 23:12












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What I think you mean is (for example), something like this $$frac{partial}{partial x}frac{partial}{partial y}w$$



This is usually denoted by $$frac{partial^2 w}{partial xpartial y}$$and is defined by $$lim_{delta xto 0}lim_{delta y to 0}left(frac{f(x+delta x,y+delta y,z)-f(x+delta x,y,z)-f(x+delta x,y,z)+f(x,y,z)}{delta xdelta y}right)$$



As an example, if we let $f(x,y,z)=x^2y^3$, then $$frac{partial^2 f}{partial xpartial y}=frac{partial }{partial x}frac{partial }{partial y}(x^2y^3)=frac{partial }{partial x}(3x^2y^2)=6xy^2$$






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    $begingroup$

    What I think you mean is (for example), something like this $$frac{partial}{partial x}frac{partial}{partial y}w$$



    This is usually denoted by $$frac{partial^2 w}{partial xpartial y}$$and is defined by $$lim_{delta xto 0}lim_{delta y to 0}left(frac{f(x+delta x,y+delta y,z)-f(x+delta x,y,z)-f(x+delta x,y,z)+f(x,y,z)}{delta xdelta y}right)$$



    As an example, if we let $f(x,y,z)=x^2y^3$, then $$frac{partial^2 f}{partial xpartial y}=frac{partial }{partial x}frac{partial }{partial y}(x^2y^3)=frac{partial }{partial x}(3x^2y^2)=6xy^2$$






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      $begingroup$

      What I think you mean is (for example), something like this $$frac{partial}{partial x}frac{partial}{partial y}w$$



      This is usually denoted by $$frac{partial^2 w}{partial xpartial y}$$and is defined by $$lim_{delta xto 0}lim_{delta y to 0}left(frac{f(x+delta x,y+delta y,z)-f(x+delta x,y,z)-f(x+delta x,y,z)+f(x,y,z)}{delta xdelta y}right)$$



      As an example, if we let $f(x,y,z)=x^2y^3$, then $$frac{partial^2 f}{partial xpartial y}=frac{partial }{partial x}frac{partial }{partial y}(x^2y^3)=frac{partial }{partial x}(3x^2y^2)=6xy^2$$






      share|cite|improve this answer









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        1





        $begingroup$

        What I think you mean is (for example), something like this $$frac{partial}{partial x}frac{partial}{partial y}w$$



        This is usually denoted by $$frac{partial^2 w}{partial xpartial y}$$and is defined by $$lim_{delta xto 0}lim_{delta y to 0}left(frac{f(x+delta x,y+delta y,z)-f(x+delta x,y,z)-f(x+delta x,y,z)+f(x,y,z)}{delta xdelta y}right)$$



        As an example, if we let $f(x,y,z)=x^2y^3$, then $$frac{partial^2 f}{partial xpartial y}=frac{partial }{partial x}frac{partial }{partial y}(x^2y^3)=frac{partial }{partial x}(3x^2y^2)=6xy^2$$






        share|cite|improve this answer









        $endgroup$



        What I think you mean is (for example), something like this $$frac{partial}{partial x}frac{partial}{partial y}w$$



        This is usually denoted by $$frac{partial^2 w}{partial xpartial y}$$and is defined by $$lim_{delta xto 0}lim_{delta y to 0}left(frac{f(x+delta x,y+delta y,z)-f(x+delta x,y,z)-f(x+delta x,y,z)+f(x,y,z)}{delta xdelta y}right)$$



        As an example, if we let $f(x,y,z)=x^2y^3$, then $$frac{partial^2 f}{partial xpartial y}=frac{partial }{partial x}frac{partial }{partial y}(x^2y^3)=frac{partial }{partial x}(3x^2y^2)=6xy^2$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 22:05









        John DoeJohn Doe

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