If Partial derivatives exist and are continuos then function is differentiable

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SUMMARY

The theorem states that if the partial derivatives of a function \( f \) exist and are continuous at a point, then the function is differentiable at that point. This conclusion is supported by the application of the Mean Value Theorem and the Newton quotient for functions of two variables. The proof can be found in advanced calculus texts, which detail the necessary conditions for differentiability in terms of continuity of partial derivatives.

PREREQUISITES
  • Understanding of partial derivatives
  • Familiarity with the Mean Value Theorem
  • Knowledge of the Newton quotient for functions of two variables
  • Basic concepts of differentiability in multivariable calculus
NEXT STEPS
  • Research the proof of the theorem regarding the continuity of partial derivatives and differentiability
  • Study the Mean Value Theorem in the context of multivariable calculus
  • Explore advanced calculus textbooks that cover differentiability conditions
  • Learn about the implications of differentiability in optimization problems
USEFUL FOR

Students of calculus, mathematicians, and educators seeking a deeper understanding of differentiability in multivariable functions.

rshalloo
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Homework Statement


Hi I'm just looking for a link to the proof of this theorem:
if the partial derivatives of function f exist and are continuous at a point then the function is differentiable there

Or even the name would be helpful
Its not a homework assignment per say, just something that our lecturer said mentioned in passing but never gave a proof of and I would like it just for the sake of completeness :P

Homework Equations



Well I think the Newton quotient is [ f(a+h,b+k) -(h)f(a,b+k)-(k)f(a+h,b) ]/Sqrt[h^2+k^2]
(for 2 variables anyway)

The Attempt at a Solution


I'm guessing that it involves the mean value theorem but I'm not entirely sure :S
 
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Any good Calculus text will have that proof- but I think it is too long and complicted to be given here.

(By the way, the phrase is "per se"- "of itself".)
 

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