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







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$endgroup$



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















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