$frac{partial{z}}{partial{y}}$ if $z$ is given implicitly












0












$begingroup$


Suppose $z$ is given implicitly as:



$$e^z-x^2y-y^2z = 0$$



Find $$frac{partial{z}}{partial{y}}$$.



I let $F(x,y) = e^z-x^2y-y^2z$. Then,



$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$



$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$



Therefore,



$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$



However, the exercise cites the answer as



$$frac{x^2+2yz}{e^z-y^2}$$



For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.






multivariable-calculus chain-rule







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
    $endgroup$
    – Dair
    Dec 10 '18 at 5:17










  • $begingroup$
    I think so too. But, is there some reasoning as to why the negative just cancels out?
    $endgroup$
    – Art
    Dec 10 '18 at 5:21
















0












$begingroup$


Suppose $z$ is given implicitly as:



$$e^z-x^2y-y^2z = 0$$



Find $$frac{partial{z}}{partial{y}}$$.



I let $F(x,y) = e^z-x^2y-y^2z$. Then,



$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$



$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$



Therefore,



$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$



However, the exercise cites the answer as



$$frac{x^2+2yz}{e^z-y^2}$$



For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.






multivariable-calculus chain-rule







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
    $endgroup$
    – Dair
    Dec 10 '18 at 5:17










  • $begingroup$
    I think so too. But, is there some reasoning as to why the negative just cancels out?
    $endgroup$
    – Art
    Dec 10 '18 at 5:21














0












0








0





$begingroup$


Suppose $z$ is given implicitly as:



$$e^z-x^2y-y^2z = 0$$



Find $$frac{partial{z}}{partial{y}}$$.



I let $F(x,y) = e^z-x^2y-y^2z$. Then,



$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$



$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$



Therefore,



$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$



However, the exercise cites the answer as



$$frac{x^2+2yz}{e^z-y^2}$$



For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.






multivariable-calculus chain-rule







share|cite|improve this question









$endgroup$




Suppose $z$ is given implicitly as:



$$e^z-x^2y-y^2z = 0$$



Find $$frac{partial{z}}{partial{y}}$$.



I let $F(x,y) = e^z-x^2y-y^2z$. Then,



$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$



$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$



Therefore,



$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$



However, the exercise cites the answer as



$$frac{x^2+2yz}{e^z-y^2}$$



For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.






multivariable-calculus chain-rule




multivariable-calculus chain-rule






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 5:09









ArtArt

32618




32618












  • $begingroup$
    I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
    $endgroup$
    – Dair
    Dec 10 '18 at 5:17










  • $begingroup$
    I think so too. But, is there some reasoning as to why the negative just cancels out?
    $endgroup$
    – Art
    Dec 10 '18 at 5:21


















  • $begingroup$
    I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
    $endgroup$
    – Dair
    Dec 10 '18 at 5:17










  • $begingroup$
    I think so too. But, is there some reasoning as to why the negative just cancels out?
    $endgroup$
    – Art
    Dec 10 '18 at 5:21
















$begingroup$
I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
$endgroup$
– Dair
Dec 10 '18 at 5:17




$begingroup$
I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
$endgroup$
– Dair
Dec 10 '18 at 5:17












$begingroup$
I think so too. But, is there some reasoning as to why the negative just cancels out?
$endgroup$
– Art
Dec 10 '18 at 5:21




$begingroup$
I think so too. But, is there some reasoning as to why the negative just cancels out?
$endgroup$
– Art
Dec 10 '18 at 5:21










2 Answers
2






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$begingroup$

Looking through my copy of Stewart's calculus we have



$$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$



and similarly



$$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$



So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.



    $e^z-x^2y-y^2z = 0$



    $z_ye^z-x^2-2yz-y^2z_y=0$



    $z_y(e^z-y^2)=x^2+2yz$



    Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$



    Added



    This proves the minus sign in the formula you used.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes









      0












      $begingroup$

      Looking through my copy of Stewart's calculus we have



      $$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$



      and similarly



      $$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$



      So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Looking through my copy of Stewart's calculus we have



        $$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$



        and similarly



        $$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$



        So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Looking through my copy of Stewart's calculus we have



          $$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$



          and similarly



          $$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$



          So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.






          share|cite|improve this answer









          $endgroup$



          Looking through my copy of Stewart's calculus we have



          $$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$



          and similarly



          $$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$



          So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 5:43









          carsandpulsarscarsandpulsars

          37817




          37817























              0












              $begingroup$

              Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.



              $e^z-x^2y-y^2z = 0$



              $z_ye^z-x^2-2yz-y^2z_y=0$



              $z_y(e^z-y^2)=x^2+2yz$



              Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$



              Added



              This proves the minus sign in the formula you used.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.



                $e^z-x^2y-y^2z = 0$



                $z_ye^z-x^2-2yz-y^2z_y=0$



                $z_y(e^z-y^2)=x^2+2yz$



                Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$



                Added



                This proves the minus sign in the formula you used.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.



                  $e^z-x^2y-y^2z = 0$



                  $z_ye^z-x^2-2yz-y^2z_y=0$



                  $z_y(e^z-y^2)=x^2+2yz$



                  Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$



                  Added



                  This proves the minus sign in the formula you used.






                  share|cite|improve this answer









                  $endgroup$



                  Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.



                  $e^z-x^2y-y^2z = 0$



                  $z_ye^z-x^2-2yz-y^2z_y=0$



                  $z_y(e^z-y^2)=x^2+2yz$



                  Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$



                  Added



                  This proves the minus sign in the formula you used.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 5:44









                  Rafa BudríaRafa Budría

                  5,9101825




                  5,9101825






























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