On an application of the chain rule.












0












$begingroup$


Let $f: mathbb{R} rightarrow mathbb{R} $ be continuously differentiable, then it's well know that



begin{align*}
f(x) & = f(y) + int_0^1 frac{partial f (y- t(x-y))}{partial t} , dt \ & = f(y) + int_0^1 f' (y- t(x-y)) (x-y) , dt
end{align*}



Let $g: mathbb{R} rightarrow mathbb{R} $ also be continuously differentiable. I perform the following derivation:



begin{align*}
g(f(x)) & = g(f(y)) + int_0^1 frac{partial g( f (y- t(x-y)))}{partial t} , dt \ & = g(f(y)) + int_0^1 g'(f (y- t(x-y) ) f' (y- t(x-y) (x-y) , dt tag{1}
end{align*}



I thought this derivation was correct but taking $g(f(x)) := f(x) / x^2$ and $f(x) := x^3$ I calculate $$g'(f(x)) = frac{f'(x) x^2 - 2x f(x)} { x^4} = frac{3x^4 - 2x^4}{x^4} =1$$
and $f'(x) = 3x^2$ and plugging this into my derivation gives



begin{align*}
x & = y + int_0^1 (y- t(x-y))^2 (x-y) , dt
end{align*}
and this is just wrong. Where is the error in my derivation of the formula in $(1)$ and how could I fix it?






real-analysis integration derivatives







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $f: mathbb{R} rightarrow mathbb{R} $ be continuously differentiable, then it's well know that



    begin{align*}
    f(x) & = f(y) + int_0^1 frac{partial f (y- t(x-y))}{partial t} , dt \ & = f(y) + int_0^1 f' (y- t(x-y)) (x-y) , dt
    end{align*}



    Let $g: mathbb{R} rightarrow mathbb{R} $ also be continuously differentiable. I perform the following derivation:



    begin{align*}
    g(f(x)) & = g(f(y)) + int_0^1 frac{partial g( f (y- t(x-y)))}{partial t} , dt \ & = g(f(y)) + int_0^1 g'(f (y- t(x-y) ) f' (y- t(x-y) (x-y) , dt tag{1}
    end{align*}



    I thought this derivation was correct but taking $g(f(x)) := f(x) / x^2$ and $f(x) := x^3$ I calculate $$g'(f(x)) = frac{f'(x) x^2 - 2x f(x)} { x^4} = frac{3x^4 - 2x^4}{x^4} =1$$
    and $f'(x) = 3x^2$ and plugging this into my derivation gives



    begin{align*}
    x & = y + int_0^1 (y- t(x-y))^2 (x-y) , dt
    end{align*}
    and this is just wrong. Where is the error in my derivation of the formula in $(1)$ and how could I fix it?






    real-analysis integration derivatives







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $f: mathbb{R} rightarrow mathbb{R} $ be continuously differentiable, then it's well know that



      begin{align*}
      f(x) & = f(y) + int_0^1 frac{partial f (y- t(x-y))}{partial t} , dt \ & = f(y) + int_0^1 f' (y- t(x-y)) (x-y) , dt
      end{align*}



      Let $g: mathbb{R} rightarrow mathbb{R} $ also be continuously differentiable. I perform the following derivation:



      begin{align*}
      g(f(x)) & = g(f(y)) + int_0^1 frac{partial g( f (y- t(x-y)))}{partial t} , dt \ & = g(f(y)) + int_0^1 g'(f (y- t(x-y) ) f' (y- t(x-y) (x-y) , dt tag{1}
      end{align*}



      I thought this derivation was correct but taking $g(f(x)) := f(x) / x^2$ and $f(x) := x^3$ I calculate $$g'(f(x)) = frac{f'(x) x^2 - 2x f(x)} { x^4} = frac{3x^4 - 2x^4}{x^4} =1$$
      and $f'(x) = 3x^2$ and plugging this into my derivation gives



      begin{align*}
      x & = y + int_0^1 (y- t(x-y))^2 (x-y) , dt
      end{align*}
      and this is just wrong. Where is the error in my derivation of the formula in $(1)$ and how could I fix it?






      real-analysis integration derivatives







      share|cite|improve this question









      $endgroup$




      Let $f: mathbb{R} rightarrow mathbb{R} $ be continuously differentiable, then it's well know that



      begin{align*}
      f(x) & = f(y) + int_0^1 frac{partial f (y- t(x-y))}{partial t} , dt \ & = f(y) + int_0^1 f' (y- t(x-y)) (x-y) , dt
      end{align*}



      Let $g: mathbb{R} rightarrow mathbb{R} $ also be continuously differentiable. I perform the following derivation:



      begin{align*}
      g(f(x)) & = g(f(y)) + int_0^1 frac{partial g( f (y- t(x-y)))}{partial t} , dt \ & = g(f(y)) + int_0^1 g'(f (y- t(x-y) ) f' (y- t(x-y) (x-y) , dt tag{1}
      end{align*}



      I thought this derivation was correct but taking $g(f(x)) := f(x) / x^2$ and $f(x) := x^3$ I calculate $$g'(f(x)) = frac{f'(x) x^2 - 2x f(x)} { x^4} = frac{3x^4 - 2x^4}{x^4} =1$$
      and $f'(x) = 3x^2$ and plugging this into my derivation gives



      begin{align*}
      x & = y + int_0^1 (y- t(x-y))^2 (x-y) , dt
      end{align*}
      and this is just wrong. Where is the error in my derivation of the formula in $(1)$ and how could I fix it?






      real-analysis integration derivatives




      real-analysis integration derivatives






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Jan 22 at 1:24









      MonoliteMonolite

      1,5502926




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          $begingroup$

          There is a mess about our $g$ function. I guess you want to say $ g(x)=x $, but if $ g(x)=x $, then $ g(f(x))= f(x) $ rather than $f(x)/{x^2}$. You need a clearer definition for $g$.






          share|cite|improve this answer











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            $begingroup$

            There is a mess about our $g$ function. I guess you want to say $ g(x)=x $, but if $ g(x)=x $, then $ g(f(x))= f(x) $ rather than $f(x)/{x^2}$. You need a clearer definition for $g$.






            share|cite|improve this answer











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              0












              $begingroup$

              There is a mess about our $g$ function. I guess you want to say $ g(x)=x $, but if $ g(x)=x $, then $ g(f(x))= f(x) $ rather than $f(x)/{x^2}$. You need a clearer definition for $g$.






              share|cite|improve this answer











              $endgroup$
















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                0





                $begingroup$

                There is a mess about our $g$ function. I guess you want to say $ g(x)=x $, but if $ g(x)=x $, then $ g(f(x))= f(x) $ rather than $f(x)/{x^2}$. You need a clearer definition for $g$.






                share|cite|improve this answer











                $endgroup$



                There is a mess about our $g$ function. I guess you want to say $ g(x)=x $, but if $ g(x)=x $, then $ g(f(x))= f(x) $ rather than $f(x)/{x^2}$. You need a clearer definition for $g$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 22 at 3:10

























                answered Jan 22 at 3:04









                GaoGao

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