derivative with product rule












1












$begingroup$


I am trying to take the derivative of this function. First, I want to take the derivative of the original function, $F(x,eta)$, with respect to $eta$:



$$F(x,eta)=frac{sqrt{2}pi}{3}g(1,eta)frac{x}{[1+C(eta)(x-1)]^3}$$



Do not worry about what $g(1,eta)$ and $C(eta)$ are, just know that they are both just functions of $eta$



So, if I want to that the derivative with respect to $eta$, this is my attempt. I was hoping someone could check the work to make sure it looks ok.



$$frac{partial F}{partial eta}=frac{sqrt{2}pi}{3}left[g'(1,eta)frac{x}{[1+C(eta)(x-1)]^3}+g(1,eta)left(frac{x}{3[1+C(eta)(x-1)]^2}*frac{1}{C'(eta)(x-1)}right)right]$$



If this looks ok, I'll try and take the derivative with respect to $x$ too.






derivatives partial-derivative







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am trying to take the derivative of this function. First, I want to take the derivative of the original function, $F(x,eta)$, with respect to $eta$:



    $$F(x,eta)=frac{sqrt{2}pi}{3}g(1,eta)frac{x}{[1+C(eta)(x-1)]^3}$$



    Do not worry about what $g(1,eta)$ and $C(eta)$ are, just know that they are both just functions of $eta$



    So, if I want to that the derivative with respect to $eta$, this is my attempt. I was hoping someone could check the work to make sure it looks ok.



    $$frac{partial F}{partial eta}=frac{sqrt{2}pi}{3}left[g'(1,eta)frac{x}{[1+C(eta)(x-1)]^3}+g(1,eta)left(frac{x}{3[1+C(eta)(x-1)]^2}*frac{1}{C'(eta)(x-1)}right)right]$$



    If this looks ok, I'll try and take the derivative with respect to $x$ too.






    derivatives partial-derivative







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to take the derivative of this function. First, I want to take the derivative of the original function, $F(x,eta)$, with respect to $eta$:



      $$F(x,eta)=frac{sqrt{2}pi}{3}g(1,eta)frac{x}{[1+C(eta)(x-1)]^3}$$



      Do not worry about what $g(1,eta)$ and $C(eta)$ are, just know that they are both just functions of $eta$



      So, if I want to that the derivative with respect to $eta$, this is my attempt. I was hoping someone could check the work to make sure it looks ok.



      $$frac{partial F}{partial eta}=frac{sqrt{2}pi}{3}left[g'(1,eta)frac{x}{[1+C(eta)(x-1)]^3}+g(1,eta)left(frac{x}{3[1+C(eta)(x-1)]^2}*frac{1}{C'(eta)(x-1)}right)right]$$



      If this looks ok, I'll try and take the derivative with respect to $x$ too.






      derivatives partial-derivative







      share|cite|improve this question









      $endgroup$




      I am trying to take the derivative of this function. First, I want to take the derivative of the original function, $F(x,eta)$, with respect to $eta$:



      $$F(x,eta)=frac{sqrt{2}pi}{3}g(1,eta)frac{x}{[1+C(eta)(x-1)]^3}$$



      Do not worry about what $g(1,eta)$ and $C(eta)$ are, just know that they are both just functions of $eta$



      So, if I want to that the derivative with respect to $eta$, this is my attempt. I was hoping someone could check the work to make sure it looks ok.



      $$frac{partial F}{partial eta}=frac{sqrt{2}pi}{3}left[g'(1,eta)frac{x}{[1+C(eta)(x-1)]^3}+g(1,eta)left(frac{x}{3[1+C(eta)(x-1)]^2}*frac{1}{C'(eta)(x-1)}right)right]$$



      If this looks ok, I'll try and take the derivative with respect to $x$ too.






      derivatives partial-derivative




      derivatives partial-derivative






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 14 at 17:30









      Jackson HartJackson Hart

      5212626




      5212626






















          1 Answer
          1






          active

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          2












          $begingroup$

          The second term has some problems, it should be something like



          $$
          frac{partial F}{partial eta} = frac{sqrt{2}pi}{3}left[g'(1,eta) frac{x}{[1 + C(eta)(x - 1)]^3} - 3g(1,eta)left(frac{x(x -1 )C'(eta)}{[1 + C(eta)(x-1)]^4}right)right]
          $$



          which just comes from



          $$
          frac{{rm d}}{{rm d}eta} frac{1}{[f(eta)]^3} = frac{{rm d}}{{rm d}eta} [f(eta)]^{-3} = -3[f(eta)]^{-4}frac{{rm d}f(eta)}{{rm d}eta} = -3frac{1}{[f(eta)]^4} frac{{rm d}f(eta)}{{rm d}eta}
          $$



          Similarly, if you take the derivative w.r.t. $x$ you should obtain



          $$
          frac{partial F}{partial x} = frac{sqrt{2}pi}{3}g(1,eta)left[frac{1}{[1 + C(eta)(x - 1)]^3} -frac{3 x C(eta)}{[1 + C(eta)(x -1)]^4}right]
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ok, so I see what you did, just curious though - What is the assumption that I made that is wrong? I just took the derivative, but just 1 over it. I guess I just want to know why my way is wrong, what is the wrong assumption that I am making?
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:44








          • 1




            $begingroup$
            @JacksonHart You are using this $$ frac{{rm d}}{{rm d}eta} frac{1}{f(eta)} = frac{1}{{rm d}f/{rm d}eta} $$ which is not true, that is the assumption that is breaking your results
            $endgroup$
            – caverac
            Jan 14 at 17:46












          • $begingroup$
            Ok great, thanks for the help! Any chance you could also show what the solution would be for the derivative with respect to x?
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:47










          • $begingroup$
            actually, I think there may be a mistake in your solution. You have $(x-1)^3$, but note that this ^3 was outside the whole thing, not just the $x-1$. Do you see what I am talking about? The cubed is outside what you refer to as $f(eta)$
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:50












          • $begingroup$
            @JacksonHart Solution updated
            $endgroup$
            – caverac
            Jan 14 at 17:51











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          1 Answer
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          2












          $begingroup$

          The second term has some problems, it should be something like



          $$
          frac{partial F}{partial eta} = frac{sqrt{2}pi}{3}left[g'(1,eta) frac{x}{[1 + C(eta)(x - 1)]^3} - 3g(1,eta)left(frac{x(x -1 )C'(eta)}{[1 + C(eta)(x-1)]^4}right)right]
          $$



          which just comes from



          $$
          frac{{rm d}}{{rm d}eta} frac{1}{[f(eta)]^3} = frac{{rm d}}{{rm d}eta} [f(eta)]^{-3} = -3[f(eta)]^{-4}frac{{rm d}f(eta)}{{rm d}eta} = -3frac{1}{[f(eta)]^4} frac{{rm d}f(eta)}{{rm d}eta}
          $$



          Similarly, if you take the derivative w.r.t. $x$ you should obtain



          $$
          frac{partial F}{partial x} = frac{sqrt{2}pi}{3}g(1,eta)left[frac{1}{[1 + C(eta)(x - 1)]^3} -frac{3 x C(eta)}{[1 + C(eta)(x -1)]^4}right]
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ok, so I see what you did, just curious though - What is the assumption that I made that is wrong? I just took the derivative, but just 1 over it. I guess I just want to know why my way is wrong, what is the wrong assumption that I am making?
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:44








          • 1




            $begingroup$
            @JacksonHart You are using this $$ frac{{rm d}}{{rm d}eta} frac{1}{f(eta)} = frac{1}{{rm d}f/{rm d}eta} $$ which is not true, that is the assumption that is breaking your results
            $endgroup$
            – caverac
            Jan 14 at 17:46












          • $begingroup$
            Ok great, thanks for the help! Any chance you could also show what the solution would be for the derivative with respect to x?
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:47










          • $begingroup$
            actually, I think there may be a mistake in your solution. You have $(x-1)^3$, but note that this ^3 was outside the whole thing, not just the $x-1$. Do you see what I am talking about? The cubed is outside what you refer to as $f(eta)$
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:50












          • $begingroup$
            @JacksonHart Solution updated
            $endgroup$
            – caverac
            Jan 14 at 17:51
















          2












          $begingroup$

          The second term has some problems, it should be something like



          $$
          frac{partial F}{partial eta} = frac{sqrt{2}pi}{3}left[g'(1,eta) frac{x}{[1 + C(eta)(x - 1)]^3} - 3g(1,eta)left(frac{x(x -1 )C'(eta)}{[1 + C(eta)(x-1)]^4}right)right]
          $$



          which just comes from



          $$
          frac{{rm d}}{{rm d}eta} frac{1}{[f(eta)]^3} = frac{{rm d}}{{rm d}eta} [f(eta)]^{-3} = -3[f(eta)]^{-4}frac{{rm d}f(eta)}{{rm d}eta} = -3frac{1}{[f(eta)]^4} frac{{rm d}f(eta)}{{rm d}eta}
          $$



          Similarly, if you take the derivative w.r.t. $x$ you should obtain



          $$
          frac{partial F}{partial x} = frac{sqrt{2}pi}{3}g(1,eta)left[frac{1}{[1 + C(eta)(x - 1)]^3} -frac{3 x C(eta)}{[1 + C(eta)(x -1)]^4}right]
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ok, so I see what you did, just curious though - What is the assumption that I made that is wrong? I just took the derivative, but just 1 over it. I guess I just want to know why my way is wrong, what is the wrong assumption that I am making?
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:44








          • 1




            $begingroup$
            @JacksonHart You are using this $$ frac{{rm d}}{{rm d}eta} frac{1}{f(eta)} = frac{1}{{rm d}f/{rm d}eta} $$ which is not true, that is the assumption that is breaking your results
            $endgroup$
            – caverac
            Jan 14 at 17:46












          • $begingroup$
            Ok great, thanks for the help! Any chance you could also show what the solution would be for the derivative with respect to x?
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:47










          • $begingroup$
            actually, I think there may be a mistake in your solution. You have $(x-1)^3$, but note that this ^3 was outside the whole thing, not just the $x-1$. Do you see what I am talking about? The cubed is outside what you refer to as $f(eta)$
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:50












          • $begingroup$
            @JacksonHart Solution updated
            $endgroup$
            – caverac
            Jan 14 at 17:51














          2












          2








          2





          $begingroup$

          The second term has some problems, it should be something like



          $$
          frac{partial F}{partial eta} = frac{sqrt{2}pi}{3}left[g'(1,eta) frac{x}{[1 + C(eta)(x - 1)]^3} - 3g(1,eta)left(frac{x(x -1 )C'(eta)}{[1 + C(eta)(x-1)]^4}right)right]
          $$



          which just comes from



          $$
          frac{{rm d}}{{rm d}eta} frac{1}{[f(eta)]^3} = frac{{rm d}}{{rm d}eta} [f(eta)]^{-3} = -3[f(eta)]^{-4}frac{{rm d}f(eta)}{{rm d}eta} = -3frac{1}{[f(eta)]^4} frac{{rm d}f(eta)}{{rm d}eta}
          $$



          Similarly, if you take the derivative w.r.t. $x$ you should obtain



          $$
          frac{partial F}{partial x} = frac{sqrt{2}pi}{3}g(1,eta)left[frac{1}{[1 + C(eta)(x - 1)]^3} -frac{3 x C(eta)}{[1 + C(eta)(x -1)]^4}right]
          $$






          share|cite|improve this answer











          $endgroup$



          The second term has some problems, it should be something like



          $$
          frac{partial F}{partial eta} = frac{sqrt{2}pi}{3}left[g'(1,eta) frac{x}{[1 + C(eta)(x - 1)]^3} - 3g(1,eta)left(frac{x(x -1 )C'(eta)}{[1 + C(eta)(x-1)]^4}right)right]
          $$



          which just comes from



          $$
          frac{{rm d}}{{rm d}eta} frac{1}{[f(eta)]^3} = frac{{rm d}}{{rm d}eta} [f(eta)]^{-3} = -3[f(eta)]^{-4}frac{{rm d}f(eta)}{{rm d}eta} = -3frac{1}{[f(eta)]^4} frac{{rm d}f(eta)}{{rm d}eta}
          $$



          Similarly, if you take the derivative w.r.t. $x$ you should obtain



          $$
          frac{partial F}{partial x} = frac{sqrt{2}pi}{3}g(1,eta)left[frac{1}{[1 + C(eta)(x - 1)]^3} -frac{3 x C(eta)}{[1 + C(eta)(x -1)]^4}right]
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 14 at 17:51

























          answered Jan 14 at 17:40









          caveraccaverac

          14.8k31130




          14.8k31130












          • $begingroup$
            Ok, so I see what you did, just curious though - What is the assumption that I made that is wrong? I just took the derivative, but just 1 over it. I guess I just want to know why my way is wrong, what is the wrong assumption that I am making?
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:44








          • 1




            $begingroup$
            @JacksonHart You are using this $$ frac{{rm d}}{{rm d}eta} frac{1}{f(eta)} = frac{1}{{rm d}f/{rm d}eta} $$ which is not true, that is the assumption that is breaking your results
            $endgroup$
            – caverac
            Jan 14 at 17:46












          • $begingroup$
            Ok great, thanks for the help! Any chance you could also show what the solution would be for the derivative with respect to x?
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:47










          • $begingroup$
            actually, I think there may be a mistake in your solution. You have $(x-1)^3$, but note that this ^3 was outside the whole thing, not just the $x-1$. Do you see what I am talking about? The cubed is outside what you refer to as $f(eta)$
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:50












          • $begingroup$
            @JacksonHart Solution updated
            $endgroup$
            – caverac
            Jan 14 at 17:51


















          • $begingroup$
            Ok, so I see what you did, just curious though - What is the assumption that I made that is wrong? I just took the derivative, but just 1 over it. I guess I just want to know why my way is wrong, what is the wrong assumption that I am making?
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:44








          • 1




            $begingroup$
            @JacksonHart You are using this $$ frac{{rm d}}{{rm d}eta} frac{1}{f(eta)} = frac{1}{{rm d}f/{rm d}eta} $$ which is not true, that is the assumption that is breaking your results
            $endgroup$
            – caverac
            Jan 14 at 17:46












          • $begingroup$
            Ok great, thanks for the help! Any chance you could also show what the solution would be for the derivative with respect to x?
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:47










          • $begingroup$
            actually, I think there may be a mistake in your solution. You have $(x-1)^3$, but note that this ^3 was outside the whole thing, not just the $x-1$. Do you see what I am talking about? The cubed is outside what you refer to as $f(eta)$
            $endgroup$
            – Jackson Hart
            Jan 14 at 17:50












          • $begingroup$
            @JacksonHart Solution updated
            $endgroup$
            – caverac
            Jan 14 at 17:51
















          $begingroup$
          Ok, so I see what you did, just curious though - What is the assumption that I made that is wrong? I just took the derivative, but just 1 over it. I guess I just want to know why my way is wrong, what is the wrong assumption that I am making?
          $endgroup$
          – Jackson Hart
          Jan 14 at 17:44






          $begingroup$
          Ok, so I see what you did, just curious though - What is the assumption that I made that is wrong? I just took the derivative, but just 1 over it. I guess I just want to know why my way is wrong, what is the wrong assumption that I am making?
          $endgroup$
          – Jackson Hart
          Jan 14 at 17:44






          1




          1




          $begingroup$
          @JacksonHart You are using this $$ frac{{rm d}}{{rm d}eta} frac{1}{f(eta)} = frac{1}{{rm d}f/{rm d}eta} $$ which is not true, that is the assumption that is breaking your results
          $endgroup$
          – caverac
          Jan 14 at 17:46






          $begingroup$
          @JacksonHart You are using this $$ frac{{rm d}}{{rm d}eta} frac{1}{f(eta)} = frac{1}{{rm d}f/{rm d}eta} $$ which is not true, that is the assumption that is breaking your results
          $endgroup$
          – caverac
          Jan 14 at 17:46














          $begingroup$
          Ok great, thanks for the help! Any chance you could also show what the solution would be for the derivative with respect to x?
          $endgroup$
          – Jackson Hart
          Jan 14 at 17:47




          $begingroup$
          Ok great, thanks for the help! Any chance you could also show what the solution would be for the derivative with respect to x?
          $endgroup$
          – Jackson Hart
          Jan 14 at 17:47












          $begingroup$
          actually, I think there may be a mistake in your solution. You have $(x-1)^3$, but note that this ^3 was outside the whole thing, not just the $x-1$. Do you see what I am talking about? The cubed is outside what you refer to as $f(eta)$
          $endgroup$
          – Jackson Hart
          Jan 14 at 17:50






          $begingroup$
          actually, I think there may be a mistake in your solution. You have $(x-1)^3$, but note that this ^3 was outside the whole thing, not just the $x-1$. Do you see what I am talking about? The cubed is outside what you refer to as $f(eta)$
          $endgroup$
          – Jackson Hart
          Jan 14 at 17:50














          $begingroup$
          @JacksonHart Solution updated
          $endgroup$
          – caverac
          Jan 14 at 17:51




          $begingroup$
          @JacksonHart Solution updated
          $endgroup$
          – caverac
          Jan 14 at 17:51


















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          張江高科駅

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          張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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