Rules for inverse functions and partial derivatives












0












$begingroup$


I have that $u(t,x)$ satisfies
$partial u/partial t + u cdot partial u/partial x = 0$



I need to show that if $x = x(t)$, then $dx/dt = u(t,x)$



So far I have



$u = -partial u/partial t cdot (partial u/partial x)^{-1}$



However, I am not sure if I can invert $partial u/partial x$, if I can cancel out the $partial u$ or what happens to the negative?



Is there another approach to solving this problem?






multivariable-calculus partial-derivative implicit-differentiation chain-rule







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I have that $u(t,x)$ satisfies
    $partial u/partial t + u cdot partial u/partial x = 0$



    I need to show that if $x = x(t)$, then $dx/dt = u(t,x)$



    So far I have



    $u = -partial u/partial t cdot (partial u/partial x)^{-1}$



    However, I am not sure if I can invert $partial u/partial x$, if I can cancel out the $partial u$ or what happens to the negative?



    Is there another approach to solving this problem?






    multivariable-calculus partial-derivative implicit-differentiation chain-rule







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have that $u(t,x)$ satisfies
      $partial u/partial t + u cdot partial u/partial x = 0$



      I need to show that if $x = x(t)$, then $dx/dt = u(t,x)$



      So far I have



      $u = -partial u/partial t cdot (partial u/partial x)^{-1}$



      However, I am not sure if I can invert $partial u/partial x$, if I can cancel out the $partial u$ or what happens to the negative?



      Is there another approach to solving this problem?






      multivariable-calculus partial-derivative implicit-differentiation chain-rule







      share|cite|improve this question











      $endgroup$




      I have that $u(t,x)$ satisfies
      $partial u/partial t + u cdot partial u/partial x = 0$



      I need to show that if $x = x(t)$, then $dx/dt = u(t,x)$



      So far I have



      $u = -partial u/partial t cdot (partial u/partial x)^{-1}$



      However, I am not sure if I can invert $partial u/partial x$, if I can cancel out the $partial u$ or what happens to the negative?



      Is there another approach to solving this problem?






      multivariable-calculus partial-derivative implicit-differentiation chain-rule




      multivariable-calculus partial-derivative implicit-differentiation chain-rule






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 23:04







      Lollipop

















      asked Dec 12 '18 at 21:54









      LollipopLollipop

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