Proving $\frac {\partial ^2 f}{\partial x\, \partial y} = \frac {\partial ^2 f}{\partial y \,\partial x}$ by the derivative theorem

Asked Jun 27 '17 at 16:15

Active Jun 27 '17 at 17:10

Viewed 327 times

If I want to prove that $\dfrac {\partial ^2 f}{\partial x\, \partial y} = \dfrac {\partial ^2 f}{\partial y \,\partial x}$, can I prove it by the definition of derivative using a double limit?

I know how to do it, but I am not sure it's enough.
The first condition is that the partial derivatives must exist and be continuous at my given point. The limit theorem says if the limit exists in some point $(x_0,y_0)$ then the examined function is continuous here. For the multivariable functions the limit exists if it doesn't matter if I calculate it first with respect to $x$ and then to $y$ and vice versa.

edited Jun 27 '17 at 17:10

Michael Hardy

asked Jun 27 '17 at 16:15

Leif

1,553
1
13
27

maybe this will help you https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Schwarz.27s_theorem – segevp Jun 27 '17 at 16:18
See also Schwarz's theorem – Karlo Jun 27 '17 at 16:18
This is my source. https://calculus.subwiki.org/wiki/Clairaut's_theorem_on_equality_of_mixed_partials – Leif Jun 27 '17 at 16:28
4

Possible duplicate of Equality of mixed partial derivatives ; See also https://math.stackexchange.com/questions/2337004/does-one-always-have-f-xy-f-yx?noredirect=1&lq=1 ; https://math.stackexchange.com/questions/29536/asymmetric-hessian-matrix?noredirect=1&lq=1 ; – Winther Jun 27 '17 at 16:36

Proving $\frac {\partial ^2 f}{\partial x\, \partial y} = \frac {\partial ^2 f}{\partial y \,\partial x}$ by the derivative theorem

0 Answers0