How do you differentiate with respect to y?












3












$begingroup$


Find the gradient of



$$z=x^y$$



I understand how to get it with respect to $x$ since $y$ is treated as a constant. But when trying to solve it with respect to $y$, why is it incorrect to implicitly differentiate and use the product rule:



$$ln(z)=ycdot ln(x)$$



$$frac{(z_y)}{z}=Bigr(ycdot frac{1}{x}Bigr)+(1cdot ln(x))$$



$$z_y=zBigr(frac{y}{x}+ln(x)Bigr)$$



$$z_y=x^yBigr(frac{y}{x}+ln(x)Bigr)$$






derivatives







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$endgroup$

















    3












    $begingroup$


    Find the gradient of



    $$z=x^y$$



    I understand how to get it with respect to $x$ since $y$ is treated as a constant. But when trying to solve it with respect to $y$, why is it incorrect to implicitly differentiate and use the product rule:



    $$ln(z)=ycdot ln(x)$$



    $$frac{(z_y)}{z}=Bigr(ycdot frac{1}{x}Bigr)+(1cdot ln(x))$$



    $$z_y=zBigr(frac{y}{x}+ln(x)Bigr)$$



    $$z_y=x^yBigr(frac{y}{x}+ln(x)Bigr)$$






    derivatives







    share|cite|improve this question









    New contributor




    Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Find the gradient of



      $$z=x^y$$



      I understand how to get it with respect to $x$ since $y$ is treated as a constant. But when trying to solve it with respect to $y$, why is it incorrect to implicitly differentiate and use the product rule:



      $$ln(z)=ycdot ln(x)$$



      $$frac{(z_y)}{z}=Bigr(ycdot frac{1}{x}Bigr)+(1cdot ln(x))$$



      $$z_y=zBigr(frac{y}{x}+ln(x)Bigr)$$



      $$z_y=x^yBigr(frac{y}{x}+ln(x)Bigr)$$






      derivatives







      share|cite|improve this question









      New contributor




      Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Find the gradient of



      $$z=x^y$$



      I understand how to get it with respect to $x$ since $y$ is treated as a constant. But when trying to solve it with respect to $y$, why is it incorrect to implicitly differentiate and use the product rule:



      $$ln(z)=ycdot ln(x)$$



      $$frac{(z_y)}{z}=Bigr(ycdot frac{1}{x}Bigr)+(1cdot ln(x))$$



      $$z_y=zBigr(frac{y}{x}+ln(x)Bigr)$$



      $$z_y=x^yBigr(frac{y}{x}+ln(x)Bigr)$$






      derivatives




      derivatives






      share|cite|improve this question









      New contributor




      Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 4 hours ago









      Xander Henderson

      14.2k103554




      14.2k103554






      New contributor




      Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 4 hours ago









      Random StudentRandom Student

      182




      182




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      Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Check out our Code of Conduct.






















          5 Answers
          5






          active

          oldest

          votes


















          1












          $begingroup$

          In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
          begin{align*}
          y = a^x
          &implies log(y) = xlog(a) \
          &implies frac{1}{y} y' = log(a) \
          &implies y' = ylog(a) = a^x log(a).
          end{align*}
          In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
          $$ z_y
          = frac{mathrm{d}}{mathrm{d}y} x^y
          = x^ylog(x).
          $$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?



            If the former then (using your notation) you get $frac{z_y}{z}=1 times ln(x)$ and so $z_y = x^y ln(x)$



            If the latter then you get $frac{z_y}{z}=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^{y-1} x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^{y-1}+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:



                $$ln(z)=ycdotln(x)$$



                $$frac{(z_y)}{z}=1cdotln(x)$$



                $$z_y=zln(x)$$



                $$z_y=x^yln(x)$$






                share|cite|improve this answer









                $endgroup$





















                  0












                  $begingroup$

                  The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,



                  $nabla z = (z_x, z_y), tag 1$



                  which is a vector. We have



                  $z = x^y, tag 2$



                  whence



                  $z_x = yx^{y - 1}, tag 3$



                  as our OP Random Student has noted. As for $z_y$, we have



                  $ln z = y ln x, tag 4$



                  whence



                  $dfrac{z_y}{z} = ln x, tag 5$



                  or



                  $z_y = z ln x = x^y ln x; tag 6$



                  thus,



                  $nabla z = (z_x, z_y) = (yx^{y - 1}, x^y ln x). tag 7$



                  In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.






                  share|cite|improve this answer











                  $endgroup$













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                    5 Answers
                    5






                    active

                    oldest

                    votes








                    5 Answers
                    5






                    active

                    oldest

                    votes









                    active

                    oldest

                    votes






                    active

                    oldest

                    votes









                    1












                    $begingroup$

                    In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
                    begin{align*}
                    y = a^x
                    &implies log(y) = xlog(a) \
                    &implies frac{1}{y} y' = log(a) \
                    &implies y' = ylog(a) = a^x log(a).
                    end{align*}
                    In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
                    $$ z_y
                    = frac{mathrm{d}}{mathrm{d}y} x^y
                    = x^ylog(x).
                    $$






                    share|cite|improve this answer









                    $endgroup$


















                      1












                      $begingroup$

                      In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
                      begin{align*}
                      y = a^x
                      &implies log(y) = xlog(a) \
                      &implies frac{1}{y} y' = log(a) \
                      &implies y' = ylog(a) = a^x log(a).
                      end{align*}
                      In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
                      $$ z_y
                      = frac{mathrm{d}}{mathrm{d}y} x^y
                      = x^ylog(x).
                      $$






                      share|cite|improve this answer









                      $endgroup$
















                        1












                        1








                        1





                        $begingroup$

                        In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
                        begin{align*}
                        y = a^x
                        &implies log(y) = xlog(a) \
                        &implies frac{1}{y} y' = log(a) \
                        &implies y' = ylog(a) = a^x log(a).
                        end{align*}
                        In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
                        $$ z_y
                        = frac{mathrm{d}}{mathrm{d}y} x^y
                        = x^ylog(x).
                        $$






                        share|cite|improve this answer









                        $endgroup$



                        In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
                        begin{align*}
                        y = a^x
                        &implies log(y) = xlog(a) \
                        &implies frac{1}{y} y' = log(a) \
                        &implies y' = ylog(a) = a^x log(a).
                        end{align*}
                        In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
                        $$ z_y
                        = frac{mathrm{d}}{mathrm{d}y} x^y
                        = x^ylog(x).
                        $$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 4 hours ago









                        Xander HendersonXander Henderson

                        14.2k103554




                        14.2k103554























                            1












                            $begingroup$

                            When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?



                            If the former then (using your notation) you get $frac{z_y}{z}=1 times ln(x)$ and so $z_y = x^y ln(x)$



                            If the latter then you get $frac{z_y}{z}=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^{y-1} x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^{y-1}+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$






                            share|cite|improve this answer









                            $endgroup$


















                              1












                              $begingroup$

                              When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?



                              If the former then (using your notation) you get $frac{z_y}{z}=1 times ln(x)$ and so $z_y = x^y ln(x)$



                              If the latter then you get $frac{z_y}{z}=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^{y-1} x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^{y-1}+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$






                              share|cite|improve this answer









                              $endgroup$
















                                1












                                1








                                1





                                $begingroup$

                                When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?



                                If the former then (using your notation) you get $frac{z_y}{z}=1 times ln(x)$ and so $z_y = x^y ln(x)$



                                If the latter then you get $frac{z_y}{z}=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^{y-1} x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^{y-1}+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$






                                share|cite|improve this answer









                                $endgroup$



                                When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?



                                If the former then (using your notation) you get $frac{z_y}{z}=1 times ln(x)$ and so $z_y = x^y ln(x)$



                                If the latter then you get $frac{z_y}{z}=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^{y-1} x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^{y-1}+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 4 hours ago









                                HenryHenry

                                98.7k476163




                                98.7k476163























                                    0












                                    $begingroup$

                                    When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      0












                                      $begingroup$

                                      When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$

                                        When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.






                                        share|cite|improve this answer









                                        $endgroup$



                                        When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 4 hours ago









                                        jmerryjmerry

                                        3,592514




                                        3,592514























                                            0












                                            $begingroup$

                                            Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:



                                            $$ln(z)=ycdotln(x)$$



                                            $$frac{(z_y)}{z}=1cdotln(x)$$



                                            $$z_y=zln(x)$$



                                            $$z_y=x^yln(x)$$






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:



                                              $$ln(z)=ycdotln(x)$$



                                              $$frac{(z_y)}{z}=1cdotln(x)$$



                                              $$z_y=zln(x)$$



                                              $$z_y=x^yln(x)$$






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:



                                                $$ln(z)=ycdotln(x)$$



                                                $$frac{(z_y)}{z}=1cdotln(x)$$



                                                $$z_y=zln(x)$$



                                                $$z_y=x^yln(x)$$






                                                share|cite|improve this answer









                                                $endgroup$



                                                Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:



                                                $$ln(z)=ycdotln(x)$$



                                                $$frac{(z_y)}{z}=1cdotln(x)$$



                                                $$z_y=zln(x)$$



                                                $$z_y=x^yln(x)$$







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered 4 hours ago









                                                Thomas ShelbyThomas Shelby

                                                2,119220




                                                2,119220























                                                    0












                                                    $begingroup$

                                                    The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,



                                                    $nabla z = (z_x, z_y), tag 1$



                                                    which is a vector. We have



                                                    $z = x^y, tag 2$



                                                    whence



                                                    $z_x = yx^{y - 1}, tag 3$



                                                    as our OP Random Student has noted. As for $z_y$, we have



                                                    $ln z = y ln x, tag 4$



                                                    whence



                                                    $dfrac{z_y}{z} = ln x, tag 5$



                                                    or



                                                    $z_y = z ln x = x^y ln x; tag 6$



                                                    thus,



                                                    $nabla z = (z_x, z_y) = (yx^{y - 1}, x^y ln x). tag 7$



                                                    In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.






                                                    share|cite|improve this answer











                                                    $endgroup$


















                                                      0












                                                      $begingroup$

                                                      The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,



                                                      $nabla z = (z_x, z_y), tag 1$



                                                      which is a vector. We have



                                                      $z = x^y, tag 2$



                                                      whence



                                                      $z_x = yx^{y - 1}, tag 3$



                                                      as our OP Random Student has noted. As for $z_y$, we have



                                                      $ln z = y ln x, tag 4$



                                                      whence



                                                      $dfrac{z_y}{z} = ln x, tag 5$



                                                      or



                                                      $z_y = z ln x = x^y ln x; tag 6$



                                                      thus,



                                                      $nabla z = (z_x, z_y) = (yx^{y - 1}, x^y ln x). tag 7$



                                                      In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.






                                                      share|cite|improve this answer











                                                      $endgroup$
















                                                        0












                                                        0








                                                        0





                                                        $begingroup$

                                                        The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,



                                                        $nabla z = (z_x, z_y), tag 1$



                                                        which is a vector. We have



                                                        $z = x^y, tag 2$



                                                        whence



                                                        $z_x = yx^{y - 1}, tag 3$



                                                        as our OP Random Student has noted. As for $z_y$, we have



                                                        $ln z = y ln x, tag 4$



                                                        whence



                                                        $dfrac{z_y}{z} = ln x, tag 5$



                                                        or



                                                        $z_y = z ln x = x^y ln x; tag 6$



                                                        thus,



                                                        $nabla z = (z_x, z_y) = (yx^{y - 1}, x^y ln x). tag 7$



                                                        In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.






                                                        share|cite|improve this answer











                                                        $endgroup$



                                                        The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,



                                                        $nabla z = (z_x, z_y), tag 1$



                                                        which is a vector. We have



                                                        $z = x^y, tag 2$



                                                        whence



                                                        $z_x = yx^{y - 1}, tag 3$



                                                        as our OP Random Student has noted. As for $z_y$, we have



                                                        $ln z = y ln x, tag 4$



                                                        whence



                                                        $dfrac{z_y}{z} = ln x, tag 5$



                                                        or



                                                        $z_y = z ln x = x^y ln x; tag 6$



                                                        thus,



                                                        $nabla z = (z_x, z_y) = (yx^{y - 1}, x^y ln x). tag 7$



                                                        In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.







                                                        share|cite|improve this answer














                                                        share|cite|improve this answer



                                                        share|cite|improve this answer








                                                        edited 2 hours ago

























                                                        answered 4 hours ago









                                                        Robert LewisRobert Lewis

                                                        44.5k22964




                                                        44.5k22964






















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                                                            Puebla de Zaragoza

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                                                            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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                                                            Musa

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                                                            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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