Does the second order differentiability always imply the first order differentiability?











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If one considers the one-variable function, we have the definition




$$f''(x):=(f'(x))'$$




So the second order derivative must be first provided the existence of the first order derivative.



But if we consider the multivariable funtion, namely $f(x,y),$ and we also have the existence $displaystyle frac{partial^2f}{partial xpartial y},$ that by definition is




$$frac{partial^2f}{partial xpartial y}:=frac{partialleft(frac{partial f}{partial x}right)}{partial y}$$




So if these imply the existence of $displaystylefrac{partial f}{partial y}$?










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    If one considers the one-variable function, we have the definition




    $$f''(x):=(f'(x))'$$




    So the second order derivative must be first provided the existence of the first order derivative.



    But if we consider the multivariable funtion, namely $f(x,y),$ and we also have the existence $displaystyle frac{partial^2f}{partial xpartial y},$ that by definition is




    $$frac{partial^2f}{partial xpartial y}:=frac{partialleft(frac{partial f}{partial x}right)}{partial y}$$




    So if these imply the existence of $displaystylefrac{partial f}{partial y}$?










    share|cite|improve this question







    New contributor




    Minh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















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      down vote

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      up vote
      0
      down vote

      favorite











      If one considers the one-variable function, we have the definition




      $$f''(x):=(f'(x))'$$




      So the second order derivative must be first provided the existence of the first order derivative.



      But if we consider the multivariable funtion, namely $f(x,y),$ and we also have the existence $displaystyle frac{partial^2f}{partial xpartial y},$ that by definition is




      $$frac{partial^2f}{partial xpartial y}:=frac{partialleft(frac{partial f}{partial x}right)}{partial y}$$




      So if these imply the existence of $displaystylefrac{partial f}{partial y}$?










      share|cite|improve this question







      New contributor




      Minh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      If one considers the one-variable function, we have the definition




      $$f''(x):=(f'(x))'$$




      So the second order derivative must be first provided the existence of the first order derivative.



      But if we consider the multivariable funtion, namely $f(x,y),$ and we also have the existence $displaystyle frac{partial^2f}{partial xpartial y},$ that by definition is




      $$frac{partial^2f}{partial xpartial y}:=frac{partialleft(frac{partial f}{partial x}right)}{partial y}$$




      So if these imply the existence of $displaystylefrac{partial f}{partial y}$?







      multivariable-calculus






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