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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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













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












                                -

                                這個網誌中的熱門文章

                                Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

                                圖片
                                Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite 1 $$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$ I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral? calculus real-analysis analysis approximation share | cite | improve this question edited Sep 11 at 5:07 asked Sep 11 at 4:48 Mclalalala 126 8 1 What is $p$? Please give more information to make it complete. – xbh Sep 11 at 5:02 @xbh already fixed it, thanks for reminding. – Mclalalala Sep 11 at 5:10 Seems that $0$ is not singular here… [I might compute wrong] – xbh Sep 11 at 5:30 @xbh can you show me how to get the close formula for the integral above? – Mclalalala Sep 11 at 5:33
                                閱讀完整內容

                                Why am i infinitely getting the same tweet with the Twitter Search API?

                                圖片
                                up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
                                閱讀完整內容

                                Carbon dioxide

                                圖片
                                "CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
                                閱讀完整內容