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 5 hours ago









      Xander Henderson

      14.2k103554




      14.2k103554






      New contributor




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      asked 5 hours ago









      Random StudentRandom Student

      182




      182




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      New contributor





      Random Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          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 5 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 5 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 5 hours ago









                                        jmerryjmerry

                                        3,602514




                                        3,602514























                                            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 5 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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                                                            CARDNET

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                                                            CARDNET は日本カードネットワークが運営するクレジット決済ネットワーク。日本カードネットワークの略称としてもCARDNETが使用されている。 事件・事故 2017年4月15日午前11時8分からクレジットカードの決済がしづらくなる障害が発生した。障害は同日午後5時18分に復旧した。原因は、複数あるCARDNETセンターの拠点のうち1つで、2重化してあるL3スイッチの片方が故障し、1系統にトラフィックが集中したことによる輻輳が発生したため。対処としては、故障したL3スイッチ交換とシステム再起動が行われた [1] 。 出典 ^ 金子寛人 ( 2017年4月17日 ). “CARDNETのクレジット決済に6時間強障害、原因はL3スイッチ故障”. 日経コンピュータ. 2018年10月18日 閲覧。 この項目は、コンピュータに関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（PJ:コンピュータ/P:コンピュータ）。 この項目は、経済に関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（ポータル 経済学、プロジェクト 経済）。 This page is only for reference, If you need detailed information, please check here
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                                                            Boot-repair Failure: Unable to locate package grub-common:i386

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                                                            0 I am on a system dual-booting Windows and Debian (usually using the latter), and a recent Windows update appears to have wreaked havoc on the Debian boot system. I managed to boot Windows again, but not Linux. Following online suggestions, I created a Live USB of Ubuntu, and booted from that. Then I have installed and run the boot-repair utility. The utility instructs me to type several lines including sudo chroot "/mnt/boot-sav/sda6" apt-get purge -y grub*-common grub-common:i386 To which I receive the error E: Unable to locate package grub-common:i386 I am given to understand the issue may have to do with having UEFI vs BIOS (I believe I have UEFI), but do not understand this well. In any case I have subsequently run the boot-repair BootInfo Summary utility which created the following file h...
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                                                            濃尾地震

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                                                            濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸...
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