Problem in evaluating logarithm derivatives












4














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
    1 hour 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.
    1 hour ago












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    1 hour 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
    36 mins ago
















4














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
    1 hour 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.
    1 hour ago












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    1 hour 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
    36 mins ago














4












4








4


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 1 hour ago









David G. Stork

9,87021232




9,87021232










asked 1 hour ago









user2192320

213




213












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












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    1 hour 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
    36 mins ago


















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












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    1 hour 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
    36 mins ago
















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




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














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




Thanks guys; this clears things up! I appreciate the fast responses.
– user2192320
1 hour 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
36 mins 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
36 mins ago










2 Answers
2






active

oldest

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5














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
    1 hour ago










  • No problem, happy to help!
    – John Doe
    36 mins ago





















2














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








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ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    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
      1 hour ago










    • No problem, happy to help!
      – John Doe
      36 mins ago


















    5














    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
      1 hour ago










    • No problem, happy to help!
      – John Doe
      36 mins ago
















    5












    5








    5






    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 1 hour ago









    John Doe

    10.2k11134




    10.2k11134












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










    • No problem, happy to help!
      – John Doe
      36 mins ago




















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










    • No problem, happy to help!
      – John Doe
      36 mins ago


















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




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












    No problem, happy to help!
    – John Doe
    36 mins ago






    No problem, happy to help!
    – John Doe
    36 mins ago













    2














    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.























      2














      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.





















        2












        2








        2






        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 1 hour ago









        ItsJustTranscendenceBro

        611




        611




        New contributor




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        New contributor





        ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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