The derivative of $ln(x)$ [closed]

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How can one prove the following by elementary means?
$$ln(x)'=frac1x$$




Say we know that
$$e^x=sum_n=0^inftyfracx^nn!.$$




sequences-and-series functions derivatives logarithms power-series




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closed as off-topic by Jyrki Lahtonen, amWhy, Xander Henderson, Taroccoesbrocco, Jendrik Stelzner Aug 10 at 3:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Xander Henderson, Taroccoesbrocco, Jendrik Stelzner
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Can you prove $(e^x)' = e^x$? Do you know a rule for derivative of an inverse function?
    – GEdgar
    Aug 9 at 18:17










  • You can get more elementary than power series. Write $f(x) =e^x$ and use the chain rule on $f circ f^-1$.
    – Randall
    Aug 9 at 18:18







  • 1




    Starting from $exp$, check $exp'=exp$, write $x=exp(ln(x))$ and differentiate both sides. (This second thing is called the inverse function theorem but you do not need to know that to carry out this proof.)
    – Ian
    Aug 9 at 18:19











  • Can you elaborate on what you consider to be "elementary" and "non-elementary"? I'm having difficulty thinking of any proof that I wouldn't consider elementary. But it seems there is some "non-elementary" proof you are trying to avoid...
    – Eric Wofsey
    Aug 9 at 18:20











  • @EricWofsey I consider e.g. the chain rule or derivative of the inverse function as non-elementary steps.
    – user122424
    Aug 9 at 18:26














up vote
0
down vote

favorite













How can one prove the following by elementary means?
$$ln(x)'=frac1x$$




Say we know that
$$e^x=sum_n=0^inftyfracx^nn!.$$




sequences-and-series functions derivatives logarithms power-series




share|cite|improve this question














closed as off-topic by Jyrki Lahtonen, amWhy, Xander Henderson, Taroccoesbrocco, Jendrik Stelzner Aug 10 at 3:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Xander Henderson, Taroccoesbrocco, Jendrik Stelzner
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Can you prove $(e^x)' = e^x$? Do you know a rule for derivative of an inverse function?
    – GEdgar
    Aug 9 at 18:17










  • You can get more elementary than power series. Write $f(x) =e^x$ and use the chain rule on $f circ f^-1$.
    – Randall
    Aug 9 at 18:18







  • 1




    Starting from $exp$, check $exp'=exp$, write $x=exp(ln(x))$ and differentiate both sides. (This second thing is called the inverse function theorem but you do not need to know that to carry out this proof.)
    – Ian
    Aug 9 at 18:19











  • Can you elaborate on what you consider to be "elementary" and "non-elementary"? I'm having difficulty thinking of any proof that I wouldn't consider elementary. But it seems there is some "non-elementary" proof you are trying to avoid...
    – Eric Wofsey
    Aug 9 at 18:20











  • @EricWofsey I consider e.g. the chain rule or derivative of the inverse function as non-elementary steps.
    – user122424
    Aug 9 at 18:26












up vote
0
down vote

favorite









up vote
0
down vote

favorite












How can one prove the following by elementary means?
$$ln(x)'=frac1x$$




Say we know that
$$e^x=sum_n=0^inftyfracx^nn!.$$




sequences-and-series functions derivatives logarithms power-series




share|cite|improve this question















How can one prove the following by elementary means?
$$ln(x)'=frac1x$$




Say we know that
$$e^x=sum_n=0^inftyfracx^nn!.$$





sequences-and-series functions derivatives logarithms power-series





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 9 at 19:12









Math Lover

12.5k21232




12.5k21232










asked Aug 9 at 18:16









user122424

9521614




9521614




closed as off-topic by Jyrki Lahtonen, amWhy, Xander Henderson, Taroccoesbrocco, Jendrik Stelzner Aug 10 at 3:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Xander Henderson, Taroccoesbrocco, Jendrik Stelzner
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Jyrki Lahtonen, amWhy, Xander Henderson, Taroccoesbrocco, Jendrik Stelzner Aug 10 at 3:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Xander Henderson, Taroccoesbrocco, Jendrik Stelzner
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Can you prove $(e^x)' = e^x$? Do you know a rule for derivative of an inverse function?
    – GEdgar
    Aug 9 at 18:17










  • You can get more elementary than power series. Write $f(x) =e^x$ and use the chain rule on $f circ f^-1$.
    – Randall
    Aug 9 at 18:18







  • 1




    Starting from $exp$, check $exp'=exp$, write $x=exp(ln(x))$ and differentiate both sides. (This second thing is called the inverse function theorem but you do not need to know that to carry out this proof.)
    – Ian
    Aug 9 at 18:19











  • Can you elaborate on what you consider to be "elementary" and "non-elementary"? I'm having difficulty thinking of any proof that I wouldn't consider elementary. But it seems there is some "non-elementary" proof you are trying to avoid...
    – Eric Wofsey
    Aug 9 at 18:20











  • @EricWofsey I consider e.g. the chain rule or derivative of the inverse function as non-elementary steps.
    – user122424
    Aug 9 at 18:26
















  • Can you prove $(e^x)' = e^x$? Do you know a rule for derivative of an inverse function?
    – GEdgar
    Aug 9 at 18:17










  • You can get more elementary than power series. Write $f(x) =e^x$ and use the chain rule on $f circ f^-1$.
    – Randall
    Aug 9 at 18:18







  • 1




    Starting from $exp$, check $exp'=exp$, write $x=exp(ln(x))$ and differentiate both sides. (This second thing is called the inverse function theorem but you do not need to know that to carry out this proof.)
    – Ian
    Aug 9 at 18:19











  • Can you elaborate on what you consider to be "elementary" and "non-elementary"? I'm having difficulty thinking of any proof that I wouldn't consider elementary. But it seems there is some "non-elementary" proof you are trying to avoid...
    – Eric Wofsey
    Aug 9 at 18:20











  • @EricWofsey I consider e.g. the chain rule or derivative of the inverse function as non-elementary steps.
    – user122424
    Aug 9 at 18:26















Can you prove $(e^x)' = e^x$? Do you know a rule for derivative of an inverse function?
– GEdgar
Aug 9 at 18:17




Can you prove $(e^x)' = e^x$? Do you know a rule for derivative of an inverse function?
– GEdgar
Aug 9 at 18:17












You can get more elementary than power series. Write $f(x) =e^x$ and use the chain rule on $f circ f^-1$.
– Randall
Aug 9 at 18:18





You can get more elementary than power series. Write $f(x) =e^x$ and use the chain rule on $f circ f^-1$.
– Randall
Aug 9 at 18:18





1




1




Starting from $exp$, check $exp'=exp$, write $x=exp(ln(x))$ and differentiate both sides. (This second thing is called the inverse function theorem but you do not need to know that to carry out this proof.)
– Ian
Aug 9 at 18:19





Starting from $exp$, check $exp'=exp$, write $x=exp(ln(x))$ and differentiate both sides. (This second thing is called the inverse function theorem but you do not need to know that to carry out this proof.)
– Ian
Aug 9 at 18:19













Can you elaborate on what you consider to be "elementary" and "non-elementary"? I'm having difficulty thinking of any proof that I wouldn't consider elementary. But it seems there is some "non-elementary" proof you are trying to avoid...
– Eric Wofsey
Aug 9 at 18:20





Can you elaborate on what you consider to be "elementary" and "non-elementary"? I'm having difficulty thinking of any proof that I wouldn't consider elementary. But it seems there is some "non-elementary" proof you are trying to avoid...
– Eric Wofsey
Aug 9 at 18:20













@EricWofsey I consider e.g. the chain rule or derivative of the inverse function as non-elementary steps.
– user122424
Aug 9 at 18:26




@EricWofsey I consider e.g. the chain rule or derivative of the inverse function as non-elementary steps.
– user122424
Aug 9 at 18:26










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










Let $f(x)=e^x$, so $f^-1(x)=ln x$. We know that $(f circ f^-1)(x)=x$. Differentiate by the chain rule to get
$$
f'(f^-1(x)) fracddxf^-1(x)=1.
$$
As $f'=f$ this boils down to
$$
x fracddxf^-1(x)=1
$$
giving what you want.






share|cite|improve this answer



























    up vote
    1
    down vote













    If we know that, thenbeginalignexp'(x)&=left(sum_n=0^inftyfracx^nn!right)'\&=sum_n=0^inftyfracx^nn!\&=exp(x).endalignTherefore,beginalignln'(x)&=frac1exp'(ln(x))\&=frac1exp(ln(x))\&=frac1x.endalign






    share|cite|improve this answer



























      up vote
      0
      down vote













      Let $f(x)=e^x$. Then:



      $$partial log (x) over partial x = partial f^-1(x) over partial x = 1 over e^log(x) = 1/x$$.






      share|cite|improve this answer



























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted










        Let $f(x)=e^x$, so $f^-1(x)=ln x$. We know that $(f circ f^-1)(x)=x$. Differentiate by the chain rule to get
        $$
        f'(f^-1(x)) fracddxf^-1(x)=1.
        $$
        As $f'=f$ this boils down to
        $$
        x fracddxf^-1(x)=1
        $$
        giving what you want.






        share|cite|improve this answer
























          up vote
          2
          down vote



          accepted










          Let $f(x)=e^x$, so $f^-1(x)=ln x$. We know that $(f circ f^-1)(x)=x$. Differentiate by the chain rule to get
          $$
          f'(f^-1(x)) fracddxf^-1(x)=1.
          $$
          As $f'=f$ this boils down to
          $$
          x fracddxf^-1(x)=1
          $$
          giving what you want.






          share|cite|improve this answer






















            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Let $f(x)=e^x$, so $f^-1(x)=ln x$. We know that $(f circ f^-1)(x)=x$. Differentiate by the chain rule to get
            $$
            f'(f^-1(x)) fracddxf^-1(x)=1.
            $$
            As $f'=f$ this boils down to
            $$
            x fracddxf^-1(x)=1
            $$
            giving what you want.






            share|cite|improve this answer












            Let $f(x)=e^x$, so $f^-1(x)=ln x$. We know that $(f circ f^-1)(x)=x$. Differentiate by the chain rule to get
            $$
            f'(f^-1(x)) fracddxf^-1(x)=1.
            $$
            As $f'=f$ this boils down to
            $$
            x fracddxf^-1(x)=1
            $$
            giving what you want.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 9 at 18:21









            Randall

            7,2471825




            7,2471825




















                up vote
                1
                down vote













                If we know that, thenbeginalignexp'(x)&=left(sum_n=0^inftyfracx^nn!right)'\&=sum_n=0^inftyfracx^nn!\&=exp(x).endalignTherefore,beginalignln'(x)&=frac1exp'(ln(x))\&=frac1exp(ln(x))\&=frac1x.endalign






                share|cite|improve this answer
























                  up vote
                  1
                  down vote













                  If we know that, thenbeginalignexp'(x)&=left(sum_n=0^inftyfracx^nn!right)'\&=sum_n=0^inftyfracx^nn!\&=exp(x).endalignTherefore,beginalignln'(x)&=frac1exp'(ln(x))\&=frac1exp(ln(x))\&=frac1x.endalign






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    If we know that, thenbeginalignexp'(x)&=left(sum_n=0^inftyfracx^nn!right)'\&=sum_n=0^inftyfracx^nn!\&=exp(x).endalignTherefore,beginalignln'(x)&=frac1exp'(ln(x))\&=frac1exp(ln(x))\&=frac1x.endalign






                    share|cite|improve this answer












                    If we know that, thenbeginalignexp'(x)&=left(sum_n=0^inftyfracx^nn!right)'\&=sum_n=0^inftyfracx^nn!\&=exp(x).endalignTherefore,beginalignln'(x)&=frac1exp'(ln(x))\&=frac1exp(ln(x))\&=frac1x.endalign







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 9 at 18:18









                    José Carlos Santos

                    115k1699177




                    115k1699177




















                        up vote
                        0
                        down vote













                        Let $f(x)=e^x$. Then:



                        $$partial log (x) over partial x = partial f^-1(x) over partial x = 1 over e^log(x) = 1/x$$.






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote













                          Let $f(x)=e^x$. Then:



                          $$partial log (x) over partial x = partial f^-1(x) over partial x = 1 over e^log(x) = 1/x$$.






                          share|cite|improve this answer






















                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Let $f(x)=e^x$. Then:



                            $$partial log (x) over partial x = partial f^-1(x) over partial x = 1 over e^log(x) = 1/x$$.






                            share|cite|improve this answer












                            Let $f(x)=e^x$. Then:



                            $$partial log (x) over partial x = partial f^-1(x) over partial x = 1 over e^log(x) = 1/x$$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 9 at 18:21









                            David G. Stork

                            7,7362929




                            7,7362929












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