Derivative of an inverse of a function [closed]












2












$begingroup$


I'm new into Calculus. Given $f:mathbb{R}rightarrow mathbb{R}$, I know that $f(0)=2$ and $f'(0)=3$. I can't understand why, given a function $g=f^{-1}$, it is true that $g'(2)=frac{1}{3}$.






calculus functions derivatives







share|cite|improve this question









$endgroup$



closed as off-topic by TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory Jan 20 at 19:19


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
    $endgroup$
    – Martin R
    Jan 20 at 13:39










  • $begingroup$
    Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
    $endgroup$
    – Eric Lippert
    Jan 20 at 17:37
















2












$begingroup$


I'm new into Calculus. Given $f:mathbb{R}rightarrow mathbb{R}$, I know that $f(0)=2$ and $f'(0)=3$. I can't understand why, given a function $g=f^{-1}$, it is true that $g'(2)=frac{1}{3}$.






calculus functions derivatives







share|cite|improve this question









$endgroup$



closed as off-topic by TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory Jan 20 at 19:19


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
    $endgroup$
    – Martin R
    Jan 20 at 13:39










  • $begingroup$
    Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
    $endgroup$
    – Eric Lippert
    Jan 20 at 17:37














2












2








2


2



$begingroup$


I'm new into Calculus. Given $f:mathbb{R}rightarrow mathbb{R}$, I know that $f(0)=2$ and $f'(0)=3$. I can't understand why, given a function $g=f^{-1}$, it is true that $g'(2)=frac{1}{3}$.






calculus functions derivatives







share|cite|improve this question









$endgroup$




I'm new into Calculus. Given $f:mathbb{R}rightarrow mathbb{R}$, I know that $f(0)=2$ and $f'(0)=3$. I can't understand why, given a function $g=f^{-1}$, it is true that $g'(2)=frac{1}{3}$.






calculus functions derivatives




calculus functions derivatives






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 20 at 13:35









Kevin MooreKevin Moore

16918




16918




closed as off-topic by TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory Jan 20 at 19:19


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory Jan 20 at 19:19


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
    $endgroup$
    – Martin R
    Jan 20 at 13:39










  • $begingroup$
    Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
    $endgroup$
    – Eric Lippert
    Jan 20 at 17:37


















  • $begingroup$
    Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
    $endgroup$
    – Martin R
    Jan 20 at 13:39










  • $begingroup$
    Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
    $endgroup$
    – Eric Lippert
    Jan 20 at 17:37
















$begingroup$
Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
$endgroup$
– Martin R
Jan 20 at 13:39




$begingroup$
Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
$endgroup$
– Martin R
Jan 20 at 13:39












$begingroup$
Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
$endgroup$
– Eric Lippert
Jan 20 at 17:37




$begingroup$
Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
$endgroup$
– Eric Lippert
Jan 20 at 17:37










3 Answers
3






active

oldest

votes


















8












$begingroup$

The idea is to write
$$
f(g(x))=x
$$
then use the chain rule
$$
f'(g(x))g'(x)=1
$$
From this you get:
$$
g'(x)=frac{1}{f'(g(x))}
$$
With your numerical example: $f'(0)=3$ and $f(0)=2Rightarrow g(2)=0$, therefore $g'(2)=frac{1}{3}$




You can have a look at Inverse_functions_and_differentiation (wiki page) to get a more detailed discussion. One important and useful property to remember is:




It follows that a function that has a continuous derivative has an
inverse in a neighbourhood of every point where the derivative is
non-zero. This need not be true if the derivative is not continuous







share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Explanation without words [oops, was that too many words?]. enter image description here






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Let's try :



      $g=f^{-1}$;



      $g(2)=f^{-1}(f(0))=0.$



      $g(f(x))= x;$



      $g'(f(x))f'(x)=1;$
      (Chain rule)



      $g'(f(x))=frac{1}{f'(x)},$ $(f'(x) not =0)$.



      $g'(f(0))= frac{1}{f'(0)};$



      $g'(2)=frac{1}{3};$






      share|cite|improve this answer











      $endgroup$




















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        8












        $begingroup$

        The idea is to write
        $$
        f(g(x))=x
        $$
        then use the chain rule
        $$
        f'(g(x))g'(x)=1
        $$
        From this you get:
        $$
        g'(x)=frac{1}{f'(g(x))}
        $$
        With your numerical example: $f'(0)=3$ and $f(0)=2Rightarrow g(2)=0$, therefore $g'(2)=frac{1}{3}$




        You can have a look at Inverse_functions_and_differentiation (wiki page) to get a more detailed discussion. One important and useful property to remember is:




        It follows that a function that has a continuous derivative has an
        inverse in a neighbourhood of every point where the derivative is
        non-zero        . This need not be true if the derivative is not continuous







        share|cite|improve this answer











        $endgroup$


















          8












          $begingroup$

          The idea is to write
          $$
          f(g(x))=x
          $$
          then use the chain rule
          $$
          f'(g(x))g'(x)=1
          $$
          From this you get:
          $$
          g'(x)=frac{1}{f'(g(x))}
          $$
          With your numerical example: $f'(0)=3$ and $f(0)=2Rightarrow g(2)=0$, therefore $g'(2)=frac{1}{3}$




          You can have a look at Inverse_functions_and_differentiation (wiki page) to get a more detailed discussion. One important and useful property to remember is:




          It follows that a function that has a continuous derivative has an
          inverse in a neighbourhood of every point where the derivative is
          non-zero          . This need not be true if the derivative is not continuous







          share|cite|improve this answer











          $endgroup$
















            8












            8








            8





            $begingroup$

            The idea is to write
            $$
            f(g(x))=x
            $$
            then use the chain rule
            $$
            f'(g(x))g'(x)=1
            $$
            From this you get:
            $$
            g'(x)=frac{1}{f'(g(x))}
            $$
            With your numerical example: $f'(0)=3$ and $f(0)=2Rightarrow g(2)=0$, therefore $g'(2)=frac{1}{3}$




            You can have a look at Inverse_functions_and_differentiation (wiki page) to get a more detailed discussion. One important and useful property to remember is:




            It follows that a function that has a continuous derivative has an
            inverse in a neighbourhood of every point where the derivative is
            non-zero            . This need not be true if the derivative is not continuous







            share|cite|improve this answer











            $endgroup$



            The idea is to write
            $$
            f(g(x))=x
            $$
            then use the chain rule
            $$
            f'(g(x))g'(x)=1
            $$
            From this you get:
            $$
            g'(x)=frac{1}{f'(g(x))}
            $$
            With your numerical example: $f'(0)=3$ and $f(0)=2Rightarrow g(2)=0$, therefore $g'(2)=frac{1}{3}$




            You can have a look at Inverse_functions_and_differentiation (wiki page) to get a more detailed discussion. One important and useful property to remember is:




            It follows that a function that has a continuous derivative has an
            inverse in a neighbourhood of every point where the derivative is
            non-zero            . This need not be true if the derivative is not continuous








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 20 at 13:55

























            answered Jan 20 at 13:47









            Picaud VincentPicaud Vincent

            1,52439




            1,52439























                3












                $begingroup$

                Explanation without words [oops, was that too many words?]. enter image description here






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Explanation without words [oops, was that too many words?]. enter image description here






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Explanation without words [oops, was that too many words?]. enter image description here






                    share|cite|improve this answer









                    $endgroup$



                    Explanation without words [oops, was that too many words?]. enter image description here







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 20 at 14:23









                    John JoyJohn Joy

                    6,32011727




                    6,32011727























                        1












                        $begingroup$

                        Let's try :



                        $g=f^{-1}$;



                        $g(2)=f^{-1}(f(0))=0.$



                        $g(f(x))= x;$



                        $g'(f(x))f'(x)=1;$
                        (Chain rule)



                        $g'(f(x))=frac{1}{f'(x)},$ $(f'(x) not =0)$.



                        $g'(f(0))= frac{1}{f'(0)};$



                        $g'(2)=frac{1}{3};$






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Let's try :



                          $g=f^{-1}$;



                          $g(2)=f^{-1}(f(0))=0.$



                          $g(f(x))= x;$



                          $g'(f(x))f'(x)=1;$
                          (Chain rule)



                          $g'(f(x))=frac{1}{f'(x)},$ $(f'(x) not =0)$.



                          $g'(f(0))= frac{1}{f'(0)};$



                          $g'(2)=frac{1}{3};$






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Let's try :



                            $g=f^{-1}$;



                            $g(2)=f^{-1}(f(0))=0.$



                            $g(f(x))= x;$



                            $g'(f(x))f'(x)=1;$
                            (Chain rule)



                            $g'(f(x))=frac{1}{f'(x)},$ $(f'(x) not =0)$.



                            $g'(f(0))= frac{1}{f'(0)};$



                            $g'(2)=frac{1}{3};$






                            share|cite|improve this answer











                            $endgroup$



                            Let's try :



                            $g=f^{-1}$;



                            $g(2)=f^{-1}(f(0))=0.$



                            $g(f(x))= x;$



                            $g'(f(x))f'(x)=1;$
                            (Chain rule)



                            $g'(f(x))=frac{1}{f'(x)},$ $(f'(x) not =0)$.



                            $g'(f(0))= frac{1}{f'(0)};$



                            $g'(2)=frac{1}{3};$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 20 at 16:13

























                            answered Jan 20 at 13:44









                            Peter SzilasPeter Szilas

                            11.3k2822




                            11.3k2822















                                -

                                Popular posts from this blog

                                Questions related to Moebius Transform of Characteristic Function of the Primes

                                Image
                                2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
                                Read more

                                List of scandals in India

                                Image
                                List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
                                Read more

                                Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                Image
                                0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
                                Read more