Problem in evaluating logarithm derivatives












8














Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?



To be more clear,



$log{xy} = log{x} + log{y}$ (property of $log$)



$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)



Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)



Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)



Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$



$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.



My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.



What's going wrong in this example?






real-analysis calculus logarithms







share|cite|improve this question
























  • Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
    – MathIsLife12
    2 days ago










  • (This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
    – Clement C.
    2 days ago












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    2 days ago










  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – John Doe
    2 days ago
















8














Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?



To be more clear,



$log{xy} = log{x} + log{y}$ (property of $log$)



$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)



Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)



Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)



Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$



$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.



My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.



What's going wrong in this example?






real-analysis calculus logarithms







share|cite|improve this question
























  • Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
    – MathIsLife12
    2 days ago










  • (This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
    – Clement C.
    2 days ago












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    2 days ago










  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – John Doe
    2 days ago














8












8








8


1





Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?



To be more clear,



$log{xy} = log{x} + log{y}$ (property of $log$)



$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)



Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)



Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)



Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$



$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.



My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.



What's going wrong in this example?






real-analysis calculus logarithms







share|cite|improve this question















Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?



To be more clear,



$log{xy} = log{x} + log{y}$ (property of $log$)



$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)



Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)



Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)



Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$



$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.



My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.



What's going wrong in this example?






real-analysis calculus logarithms




real-analysis calculus logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









David G. Stork

9,98021232




9,98021232










asked 2 days ago









user2192320user2192320

473




473












  • Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
    – MathIsLife12
    2 days ago










  • (This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
    – Clement C.
    2 days ago












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    2 days ago










  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – John Doe
    2 days ago


















  • Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
    – MathIsLife12
    2 days ago










  • (This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
    – Clement C.
    2 days ago












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    2 days ago










  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – John Doe
    2 days ago
















Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
– MathIsLife12
2 days ago




Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
– MathIsLife12
2 days ago












(This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
– Clement C.
2 days ago






(This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
– Clement C.
2 days ago














Thanks guys; this clears things up! I appreciate the fast responses.
– user2192320
2 days ago




Thanks guys; this clears things up! I appreciate the fast responses.
– user2192320
2 days ago












After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– John Doe
2 days ago




After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– John Doe
2 days ago










2 Answers
2






active

oldest

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21














Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).



Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
So there are no inconsistencies.






share|cite|improve this answer





















  • Thank you for the quick response; this is my favorite answer posted so far.
    – user2192320
    2 days ago










  • No problem, happy to help!
    – John Doe
    2 days ago








  • 2




    @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
    – Ruslan
    2 days ago



















10














The issue is that your differential operator $D$ does not behave in the way you think it does!
$$
D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
$$
At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.






share|cite|improve this answer








New contributor




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


















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    21














    Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).



    Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
    So there are no inconsistencies.






    share|cite|improve this answer





















    • Thank you for the quick response; this is my favorite answer posted so far.
      – user2192320
      2 days ago










    • No problem, happy to help!
      – John Doe
      2 days ago








    • 2




      @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
      – Ruslan
      2 days ago
















    21














    Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).



    Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
    So there are no inconsistencies.






    share|cite|improve this answer





















    • Thank you for the quick response; this is my favorite answer posted so far.
      – user2192320
      2 days ago










    • No problem, happy to help!
      – John Doe
      2 days ago








    • 2




      @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
      – Ruslan
      2 days ago














    21












    21








    21






    Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).



    Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
    So there are no inconsistencies.






    share|cite|improve this answer












    Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).



    Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
    So there are no inconsistencies.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 days ago









    John DoeJohn Doe

    10.6k11237




    10.6k11237












    • Thank you for the quick response; this is my favorite answer posted so far.
      – user2192320
      2 days ago










    • No problem, happy to help!
      – John Doe
      2 days ago








    • 2




      @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
      – Ruslan
      2 days ago


















    • Thank you for the quick response; this is my favorite answer posted so far.
      – user2192320
      2 days ago










    • No problem, happy to help!
      – John Doe
      2 days ago








    • 2




      @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
      – Ruslan
      2 days ago
















    Thank you for the quick response; this is my favorite answer posted so far.
    – user2192320
    2 days ago




    Thank you for the quick response; this is my favorite answer posted so far.
    – user2192320
    2 days ago












    No problem, happy to help!
    – John Doe
    2 days ago






    No problem, happy to help!
    – John Doe
    2 days ago






    2




    2




    @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
    – Ruslan
    2 days ago




    @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
    – Ruslan
    2 days ago











    10














    The issue is that your differential operator $D$ does not behave in the way you think it does!
    $$
    D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
    $$
    At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.






    share|cite|improve this answer








    New contributor




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























      10














      The issue is that your differential operator $D$ does not behave in the way you think it does!
      $$
      D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
      $$
      At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.






      share|cite|improve this answer








      New contributor




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





















        10












        10








        10






        The issue is that your differential operator $D$ does not behave in the way you think it does!
        $$
        D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
        $$
        At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.






        share|cite|improve this answer








        New contributor




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









        The issue is that your differential operator $D$ does not behave in the way you think it does!
        $$
        D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
        $$
        At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.







        share|cite|improve this answer








        New contributor




        ItsJustTranscendenceBro 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 answer



        share|cite|improve this answer






        New contributor




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









        answered 2 days ago









        ItsJustTranscendenceBroItsJustTranscendenceBro

        1712




        1712




        New contributor




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





        New contributor





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






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






























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            8-я гвардейская общевойсковая ордена Ленина армия Награды: Войска: РККА → Советская армия → СВ России Род войск: Формирование: 1943 Расформирование (преобразование): 1992 Предшественник: 62-я армия (1942—43) → 8-я гвардейская общевойсковая армия (1943—92) → 8-й гвардейский армейский корпус (1992—98) Боевой путь Изюм-Барвенковская операция Донбасская операция Запорожская операция Битва за Днепр Днепропетровская операция Березнеговато-Снигирёвская операция Одесская операция Люблин-Брестская операция Варшавско-Познанская операция Восточно-Померанская операция Берлинская наступательная операция У этого термина существуют и другие значения, см. 8-я армия. Фотография Гвардейского Красного Знамени (лицевая сторона) 8-й гвардейской армии , образца 1943 года. Фотография Гвардейского Красного Знамени (оборотная сторона) 8-й гвардейской армии , образца 1943 года. 8-я гвардейская общевойсковая ордена Ленина армия  — гвардейское фор...
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            Алькесар

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            Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
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