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Implicit function derivative: Simplification



The Next CEO of Stack OverflowFinding partial derivative plug in a value?Partial Derivative of f(x,y) = x^yPartial derivativesImplicit function theorem in $Bbb R^2$Clarification on derivative of multi-variable function.Derivative of function $ y=[e^x^2-cot(ln(sqrt x+frac 1x))]^sec(frac1x) $Implicit Differentiation Given functions3d function and partial derivatviespartial derivatives and implicit differentiationDerivative of function and simplification










0












$begingroup$


Simplify the following system of inequities: $$fracd^2ydx^2<fracdyxdx$$
$$fracd^2xdy^2<fracdxydy$$



Known that $F(x,y)=0$. $F$ is monotonic.




I got something like this:



$$fracF_yyF_y-fracF_xyF_x<1/x$$
$$fracF_xxF_x-fracF_xyF_y<1/y$$



Not sure if it is correct or it could be further simplified.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    Simplify the following system of inequities: $$fracd^2ydx^2<fracdyxdx$$
    $$fracd^2xdy^2<fracdxydy$$



    Known that $F(x,y)=0$. $F$ is monotonic.




    I got something like this:



    $$fracF_yyF_y-fracF_xyF_x<1/x$$
    $$fracF_xxF_x-fracF_xyF_y<1/y$$



    Not sure if it is correct or it could be further simplified.










    share|cite|improve this question











    $endgroup$














      0












      0








      0


      1



      $begingroup$


      Simplify the following system of inequities: $$fracd^2ydx^2<fracdyxdx$$
      $$fracd^2xdy^2<fracdxydy$$



      Known that $F(x,y)=0$. $F$ is monotonic.




      I got something like this:



      $$fracF_yyF_y-fracF_xyF_x<1/x$$
      $$fracF_xxF_x-fracF_xyF_y<1/y$$



      Not sure if it is correct or it could be further simplified.










      share|cite|improve this question











      $endgroup$




      Simplify the following system of inequities: $$fracd^2ydx^2<fracdyxdx$$
      $$fracd^2xdy^2<fracdxydy$$



      Known that $F(x,y)=0$. $F$ is monotonic.




      I got something like this:



      $$fracF_yyF_y-fracF_xyF_x<1/x$$
      $$fracF_xxF_x-fracF_xyF_y<1/y$$



      Not sure if it is correct or it could be further simplified.







      real-analysis functions derivatives inequality partial-derivative






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 20 at 20:19







      High GPA

















      asked Mar 20 at 10:59









      High GPAHigh GPA

      1,008421




      1,008421




















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