Describing curve $C: f(x,y)=0$ as graph of a function $x=g(y)$ near $(1,1)$ with $f(x,y)=x^3+x^2 ln(y)-y$

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$f(x,y)=x^3+x^2 ln(y)-y$.



How can I show that in a neighborhood of $x_0=(1,1)$ the curve $C:f(x,y)=0$ can be locally described as graph of a function $x=g(y)$? I also need to determine $g'(1)$ but this should be doable once I know what $g(y)$ is.



Thanks in advance!




calculus real-analysis analysis functions




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    Do you know the Implicit Function Theorem?
    – Julián Aguirre
    Aug 29 at 9:20















up vote
0
down vote

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$f(x,y)=x^3+x^2 ln(y)-y$.



How can I show that in a neighborhood of $x_0=(1,1)$ the curve $C:f(x,y)=0$ can be locally described as graph of a function $x=g(y)$? I also need to determine $g'(1)$ but this should be doable once I know what $g(y)$ is.



Thanks in advance!




calculus real-analysis analysis functions




share|cite|improve this question
















  • 1




    Do you know the Implicit Function Theorem?
    – Julián Aguirre
    Aug 29 at 9:20













up vote
0
down vote

favorite









up vote
0
down vote

favorite











$f(x,y)=x^3+x^2 ln(y)-y$.



How can I show that in a neighborhood of $x_0=(1,1)$ the curve $C:f(x,y)=0$ can be locally described as graph of a function $x=g(y)$? I also need to determine $g'(1)$ but this should be doable once I know what $g(y)$ is.



Thanks in advance!




calculus real-analysis analysis functions




share|cite|improve this question












$f(x,y)=x^3+x^2 ln(y)-y$.



How can I show that in a neighborhood of $x_0=(1,1)$ the curve $C:f(x,y)=0$ can be locally described as graph of a function $x=g(y)$? I also need to determine $g'(1)$ but this should be doable once I know what $g(y)$ is.



Thanks in advance!





calculus real-analysis analysis functions





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 29 at 9:16









user586087

635




635







  • 1




    Do you know the Implicit Function Theorem?
    – Julián Aguirre
    Aug 29 at 9:20













  • 1




    Do you know the Implicit Function Theorem?
    – Julián Aguirre
    Aug 29 at 9:20








1




1




Do you know the Implicit Function Theorem?
– Julián Aguirre
Aug 29 at 9:20





Do you know the Implicit Function Theorem?
– Julián Aguirre
Aug 29 at 9:20











1 Answer
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We have $f(x_0)=0$ and $f_x(x_0) ne 0$. Hence the existence of a function $x=g(y)$ is garanteed by the Implicit Function Theorem.



Furthermore we have



$0=f(g(y),y)$ in a neiborhood $U$ of $1$. Thus, by the chain rule:



$(*) quad 0=f_x(g(y),y)g'(y)+f_y(g(y),y)$ for all $y in U$.



Since $g(1)=1$ you can compute $g'(1)$ with $(*)$.






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    up vote
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    down vote













    We have $f(x_0)=0$ and $f_x(x_0) ne 0$. Hence the existence of a function $x=g(y)$ is garanteed by the Implicit Function Theorem.



    Furthermore we have



    $0=f(g(y),y)$ in a neiborhood $U$ of $1$. Thus, by the chain rule:



    $(*) quad 0=f_x(g(y),y)g'(y)+f_y(g(y),y)$ for all $y in U$.



    Since $g(1)=1$ you can compute $g'(1)$ with $(*)$.






    share|cite|improve this answer
























      up vote
      0
      down vote













      We have $f(x_0)=0$ and $f_x(x_0) ne 0$. Hence the existence of a function $x=g(y)$ is garanteed by the Implicit Function Theorem.



      Furthermore we have



      $0=f(g(y),y)$ in a neiborhood $U$ of $1$. Thus, by the chain rule:



      $(*) quad 0=f_x(g(y),y)g'(y)+f_y(g(y),y)$ for all $y in U$.



      Since $g(1)=1$ you can compute $g'(1)$ with $(*)$.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        We have $f(x_0)=0$ and $f_x(x_0) ne 0$. Hence the existence of a function $x=g(y)$ is garanteed by the Implicit Function Theorem.



        Furthermore we have



        $0=f(g(y),y)$ in a neiborhood $U$ of $1$. Thus, by the chain rule:



        $(*) quad 0=f_x(g(y),y)g'(y)+f_y(g(y),y)$ for all $y in U$.



        Since $g(1)=1$ you can compute $g'(1)$ with $(*)$.






        share|cite|improve this answer












        We have $f(x_0)=0$ and $f_x(x_0) ne 0$. Hence the existence of a function $x=g(y)$ is garanteed by the Implicit Function Theorem.



        Furthermore we have



        $0=f(g(y),y)$ in a neiborhood $U$ of $1$. Thus, by the chain rule:



        $(*) quad 0=f_x(g(y),y)g'(y)+f_y(g(y),y)$ for all $y in U$.



        Since $g(1)=1$ you can compute $g'(1)$ with $(*)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 29 at 9:28









        Fred

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